This deck distinguishes absolute from relative extrema, then covers the Extreme Value Theorem and its hypotheses, critical points (including those where the derivative is undefined), Fermat's theorem, and the closed-interval method. It goes on to Rolle's Theorem, the Mean Value Theorem and its consequences, and the First Derivative Test. It targets the classic traps: missing critical points where the derivative does not exist, skipping the endpoints, applying the EVT or the MVT on an open interval or across a discontinuity, and assuming that every critical point is a maximum or a minimum.
Subject: Calculus I · 134 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 14
Where a function peaks, why a peak forces a flat tangent, and the theorem that turns average change into instantaneous change.
Objectives
By the end of this deck you will be able to:
Warm-up
Discussion prompt
Before we open Extreme Values, Rolle's Theorem, and the MVT: without looking back, what was the main idea of Indeterminate Forms and L'Hopital's Rule, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck explains what makes a limit form indeterminate rather than merely undefined, lists the seven indeterminate forms, and applies L'Hopital's Rule with its hypotheses checked every single time. It targets the four classic errors: applying the rule to a determinate form, using the quotient rule instead of differentiating the top and the bottom separately, skipping the re-check before a second application, and forgetting to exponentiate at the end of a log-then-limit problem.
Section
Section 1
Concept
An absolute (or global) maximum is the single largest output the function ever produces on the interval you care about.
absolute maximum — A number c in the domain D is where f has an absolute maximum if f(c) is greater than or equal to f(x) for every x in D. The absolute maximum VALUE is f(c); the LOCATION is c.
\[ f(c) \ge f(x) \quad \text{for all } x \in D \]
Swap the inequality and you have the absolute minimum. Notice the definition compares against the entire domain, not just nearby points.
Counterexample
Discussion prompt
An absolute (or global) maximum is the single largest output the function ever produces on the interval you care about.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Swap the inequality and you have the absolute minimum. Notice the definition compares against the entire domain, not just nearby points.
Concept
A relative (or local) maximum only has to beat its immediate neighbours. It is a summit, not necessarily the tallest summit.
relative maximum — f has a relative maximum at c if there is some open interval around c on which f(c) is the largest value. The comparison is local: only points near c count.
\[ f(c) \ge f(x) \quad \text{for all } x \text{ in some open interval containing } c \]
Every absolute extremum that sits strictly inside the interval is automatically a relative extremum too. The reverse is not true.
Analogy
Discussion prompt
Explain A relative extremum is only the best value nearby by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A relative (or local) maximum only has to beat its immediate neighbours. It is a summit, not necessarily the tallest summit.
Intuition
Imagine hiking a ridge line from one trailhead to the other. Every time the trail stops climbing and starts descending, you are standing on a local summit.
Only one of those summits is the highest point of the whole hike. That one is the absolute maximum.
A local summit is a question you answer by looking a few steps left and right. The absolute maximum is a question you answer by comparing every summit and both trailheads.
Explain it
Discussion prompt
Explain The mountain-range picture to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Imagine hiking a ridge line from one trailhead to the other. Every time the trail stops climbing and starts descending, you are standing on a local summit.
Concept
The trailheads count. On a closed interval, an absolute maximum is allowed to occur at an endpoint, where the function simply had no room to keep rising.
By the usual convention a relative extremum requires an open interval around the point, so endpoints are never relative extrema. They are still candidates for the absolute ones.
That single sentence is why the closed-interval method later on tests endpoints separately.
Picture it
Animation
Shows: Endpoints can hold an absolute extremum — a rendered Manim animation.
Rendered with Manim.
Takeaway: The largest value on this interval is at the right endpoint, not a critical point.
Picture it
Figure (svg): A curve on a closed interval starting at height 2, rising to a local maximum of 5, falling to a local minimum of 1, then rising to 7 at the right endpoint.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Here is a continuous function on a closed interval. Find every relative extremum and both absolute extrema.
Worked example
Here is a continuous function on a closed interval. Find every relative extremum and both absolute extrema.
Figure (svg): A curve on a closed interval starting at height 2, rising to a local maximum of 5, falling to a local minimum of 1, then rising to 7 at the right endpoint.
Locate the interior turning points
Why: The curve climbs then falls at one input, and falls then climbs at another. Those two places are the only interior summits and valleys.
Name the relative extrema
Why: There is a relative maximum where the output is 5 and a relative minimum where the output is 1. Both sit strictly inside the interval, so an open interval fits around each.
\[ \text{rel. max } f(1)=5, \qquad \text{rel. min } f(3)=1 \]
Collect the endpoint values too
Why: Endpoints are not relative extrema, but they are candidates for the absolute ones, so their heights go on the list.
\[ f(0)=2, \qquad f(6)=7 \]
Compare all four heights and pick the largest and smallest
Why: Absolute extrema are decided by comparing numbers, not by looking at shape. The list is 2, 5, 1, 7.
\[ \text{abs. max } f(6)=7, \qquad \text{abs. min } f(3)=1 \]
Verify by scanning the graph left to right
Why: No point on the curve rises above height 7, and the curve reaches 7 only at the right endpoint. No point dips below height 1, and it reaches 1 only at the interior valley. Both answers check out, and note the absolute maximum is NOT a relative maximum because it sits at an endpoint.
Notation
Annotate
From Worked example: reading extrema off a graph — read this one piece at a time. What is each part doing?
On: \( \text{rel. max } f(1)=5, \qquad \text{rel. min } f(3)=1 \)
Concept
Does a maximum even have to exist? Not always. The Extreme Value Theorem tells you exactly when it is guaranteed.
Extreme Value Theorem (EVT) — If f is continuous on a closed, bounded interval [a, b], then f attains an absolute maximum value and an absolute minimum value somewhere on [a, b].
\[ f \text{ continuous on } [a,b] \;\Longrightarrow\; \exists\, c,d \in [a,b] \text{ with } f(d) \le f(x) \le f(c)\ \ \forall x \in [a,b] \]
This is an existence theorem. It promises the extrema are there; it does not tell you where they are. Finding them is the next section's job.
Picture it
Animation
Shows: The Extreme Value Theorem — a rendered Manim animation.
Rendered with Manim.
Takeaway: Drop closed, or drop continuous, and the guarantee evaporates.
Intuition
Picture drawing the graph without lifting your pencil, from a definite starting point to a definite stopping point.
Closed means both ends are actually included, so the pencil has a real last point. Bounded means the interval has finite length, so the pencil cannot run off forever.
Continuous means no jumps and no holes, so the function cannot sneak up toward a height and then teleport past it without ever landing there.
Drop either hypothesis and the function can approach a best value forever without ever reaching it. That is the whole failure mode.
Picture it
Animation
Shows: Both hypotheses are load-bearing — a rendered Manim animation.
Rendered with Manim.
Takeaway: Drop either condition and the guarantee is gone.
Concept
Take the identity function on an open interval. It is perfectly continuous, but the interval is missing its endpoints.
\[ f(x) = x \quad \text{on} \quad (0,1) \]
The outputs climb toward 1 but never reach it, because the input 1 is not in the domain. Pick any candidate for the maximum and there is always a bigger one closer to the right end.
\[ \text{no absolute maximum, no absolute minimum on } (0,1) \]
Continuity alone was not enough. The interval has to be closed.
Concept
Now keep the interval closed but break continuity at one interior point.
\[ f(x) = \begin{cases} \dfrac{1}{x^{2}}, & x \ne 0 \\[6pt] 0, & x = 0 \end{cases} \quad \text{on} \quad [-1,1] \]
The interval is closed and bounded, but the function blows up near the origin, so the outputs are unbounded above.
\[ \lim_{x \to 0} \frac{1}{x^{2}} = \infty \quad \Longrightarrow \quad \text{no absolute maximum} \]
One bad point is enough to void the guarantee. Both hypotheses must hold.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The function is continuous, so the Extreme Value Theorem gives me a maximum and a minimum.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Reasoning from the endpoints as if they belonged to the interval.
Check the hypotheses first: continuous and on a closed, bounded interval.
Why: Reasoning from the endpoints as if they belonged to the interval. But f never actually outputs 9, because the inputs 3 and negative 3 are not in the domain.
Trap
The function is continuous, so the Extreme Value Theorem gives me a maximum and a minimum.
\[ f(x) = x^{2} \quad \text{on} \quad (-3, 3) \]
Report an absolute maximum value of 9
Why: Reasoning from the endpoints as if they belonged to the interval. But f never actually outputs 9, because the inputs 3 and negative 3 are not in the domain.
\[ f(x) < 9 \text{ for every } x \in (-3,3) \;\Rightarrow\; \text{9 is never attained} \]
Check the hypotheses first: continuous and on a closed, bounded interval.
\[ f(x) = x^{2} \quad \text{on} \quad (-3, 3) \]
The interval is open, so the EVT does not apply
Why: The theorem is silent here. Silent does not mean no extrema exist, so examine the function directly instead of quoting a theorem.
Examine directly: there is an absolute minimum but no absolute maximum
Why: The value 0 at the input 0 is attained and is the smallest output. Above it the outputs climb toward 9 without reaching it, so there is no absolute maximum.
\[ \text{abs. min } f(0)=0; \quad \text{no abs. max} \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
The theorem is silent here. Silent does not mean no extrema exist, so examine the function directly instead of quoting a theorem.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Reasoning from the endpoints as if they belonged to the interval. But f never actually outputs 9, because the inputs 3 and negative 3 are not in the domain.
Section
Section 2
Intuition
Stand at an interior high point of a smooth curve. Walk a tiny step right: you go down or stay level. Walk a tiny step left: same.
If the tangent line sloped upward, then stepping right would take you higher, and you were not at a high point after all. If it sloped downward, stepping left would take you higher.
The only slope left is zero. A smooth interior peak or valley has a horizontal tangent.
Concept
That argument is a real theorem, and it is the engine behind every optimization technique in this course.
Fermat's Theorem — If f has a relative extremum at an interior point c AND f is differentiable at c, then the derivative at c is zero.
\[ f \text{ has a rel. extremum at } c \ \text{ and } \ f'(c) \text{ exists} \;\Longrightarrow\; f'(c) = 0 \]
Read the arrow carefully. It runs one way: extremum forces flat tangent. It does not say a flat tangent forces an extremum.
Picture it
Animation
Shows: The argument that a non-zero slope always offers a way to improve.
A non-zero slope is always an escape route.
Takeaway: If the slope were positive you could step right and go higher; negative, step left. Either way it was not a maximum, so the slope must be zero.
Concept
Fermat's theorem needed the derivative to exist. So relative extrema hide in exactly two kinds of places: where the derivative is zero, and where it does not exist at all.
critical point (critical number) — An interior point c of the domain of f where either the derivative is zero, or the derivative fails to exist. The function itself must be defined at c.
\[ c \text{ is critical} \iff f'(c) = 0 \ \ \text{or} \ \ f'(c) \text{ does not exist} \]
The second half of that definition is the single most forgotten line in this chapter. Circle it.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of relative maximum, Extreme Value Theorem (EVT), Fermat's Theorem, critical point (critical number) as Extreme Values, Rolle's Theorem, and the MVT uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: A cubic with two visible flat points.
Two flat spots: one a peak, one a valley.
Takeaway: Critical points are where the derivative is zero or fails to exist. They are candidates for extrema, not guarantees of them.
Ranking
Put in order
Put the moves of Worked example: critical points of a cubic into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Critical points are defined by the derivative, so the derivative is the first thing to build.
Worked example
Find every critical point.
\[ f(x) = x^{3} - 3x^{2} - 9x + 5 \]
Differentiate term by term with the power rule
Why: Critical points are defined by the derivative, so the derivative is the first thing to build.
\[ f'(x) = 3x^{2} - 6x - 9 \]
Factor out the common 3, then factor the quadratic
Why: A factored derivative both solves the equation and, later, hands you the sign chart for free.
\[ f'(x) = 3\left(x^{2} - 2x - 3\right) = 3(x-3)(x+1) \]
Set each factor equal to zero
Why: A product is zero exactly when one of its factors is zero.
\[ x - 3 = 0 \ \Rightarrow\ x = 3, \qquad x + 1 = 0 \ \Rightarrow\ x = -1 \]
Ask whether the derivative is ever undefined
Why: A polynomial derivative is defined for every real number, so there are no extra critical points from the undefined case here.
\[ \text{critical numbers: } x = -1 \ \text{ and } \ x = 3 \]
Verify by substituting both numbers back into the derivative
Why: Substituting negative 1 gives three times one plus six minus nine, which is zero. Substituting 3 gives twenty-seven minus eighteen minus nine, which is also zero. Both check.
\[ f'(-1) = 3 + 6 - 9 = 0, \qquad f'(3) = 27 - 18 - 9 = 0 \]
Concept
A function can turn around at a point where it has no tangent line at all. The graph gets a corner or a cusp, and the derivative simply does not exist there.
The cleanest example is a two-thirds power. Its graph comes down to a sharp point and goes right back up.
Figure (svg): A cusp: two branches meeting at a sharp downward point with vertical tangents on both sides.
\[ f(x) = x^{2/3}, \qquad f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \]
The derivative is never zero, yet the function obviously bottoms out at the origin. That minimum lives entirely in the undefined case.
Step zero
Discussion prompt
Worked example: a critical point the derivative hides — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate with the power rule on each fractional exponent
Answer:
Worked example
Find every critical number.
\[ f(x) = x^{5/3} - 5x^{2/3} \]
Differentiate with the power rule on each fractional exponent
Why: Subtract one from each exponent: five-thirds minus one is two-thirds, and two-thirds minus one is negative one-third.
\[ f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} \]
Factor out the most negative power
Why: Pulling out the negative one-third power turns the difference into a product, which exposes both the zero and the undefined behaviour at once.
\[ f'(x) = \frac{5}{3}x^{-1/3}\left(x - 2\right) = \frac{5(x-2)}{3\sqrt[3]{x}} \]
Set the numerator to zero
Why: A fraction is zero exactly when its numerator is zero and its denominator is not.
\[ 5(x-2) = 0 \ \Longrightarrow \ x = 2 \]
Set the denominator to zero to find where the derivative dies
Why: The cube root of zero is zero, so the derivative is undefined at the origin. The original function IS defined there, so it counts as a critical number.
\[ 3\sqrt[3]{x} = 0 \ \Longrightarrow \ x = 0, \qquad f(0) = 0 \ \text{is defined} \]
\[ \text{critical numbers: } x = 0 \ \text{ and } \ x = 2 \]
Verify the factored derivative against the unfactored one
Why: At the input 8, the unfactored form gives five-thirds times four minus ten-thirds times one-half, which is twenty-thirds minus five-thirds, equal to 5. The factored form gives five-thirds times one-half times six, also 5. The factoring is correct, so the two critical numbers stand.
\[ f'(8) = \tfrac{5}{3}(4) - \tfrac{10}{3}\!\left(\tfrac12\right) = \tfrac{20}{3} - \tfrac{5}{3} = 5 \;=\; \frac{5(8-2)}{3\sqrt[3]{8}} = \frac{30}{6} = 5 \]
Picture it
Animation
Shows: Each line of the worked example "a critical point the derivative hides", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the input 8, the unfactored form gives five-thirds times four minus ten-thirds times one-half, which is twenty-thirds minus five-thirds, equal to 5. The factored form gives five-thirds times one-half times six, also 5. The factoring is correct, so the two critical numbers stand.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Find the critical numbers of this function.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The habit from polynomial problems: critical points are where the derivative is zero, so solve that one equation.
Find the critical numbers of this function.
Why: The habit from polynomial problems: critical points are where the derivative is zero, so solve that one equation.
Trap
Find the critical numbers of this function.
\[ g(x) = (x-4)^{2/3} \]
Differentiate, set equal to zero, and solve
Why: The habit from polynomial problems: critical points are where the derivative is zero, so solve that one equation.
\[ g'(x) = \frac{2}{3}(x-4)^{-1/3} = \frac{2}{3\sqrt[3]{x-4}} \]
Conclude there are no critical numbers
Why: The numerator is the constant 2, which is never zero, so the equation has no solution. The search stops here - and the actual minimum of the function is missed completely.
\[ \tfrac{2}{3} \ne 0 \ \Rightarrow\ \text{``no critical numbers''} \quad \text{(wrong)} \]
Find the critical numbers of this function.
\[ g(x) = (x-4)^{2/3} \]
Differentiate, then check BOTH cases
Why: Critical numbers come from the derivative being zero or from the derivative not existing. Two questions, not one.
\[ g'(x) = \frac{2}{3\sqrt[3]{x-4}} \]
Case 2: the denominator vanishes at the input 4
Why: The cube root of zero is zero, so the derivative is undefined at 4. The function itself is defined there, with output zero, so 4 is a genuine critical number.
\[ g(4) = 0 \ \text{ defined}, \quad g'(4) \ \text{ undefined} \ \Rightarrow\ x = 4 \text{ is critical} \]
That critical number is the absolute minimum
Why: A two-thirds power is never negative, and it equals zero only when the base is zero. So the cusp really is the lowest point on the graph.
\[ \text{abs. min } g(4) = 0 \]
Translation
\( g'(x) = \frac{2}{3}(x-4)^{-1/3} = \frac{2}{3\sqrt[3]{x-4}} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Critical points are candidates, not winners. Some of them are peaks, some are valleys, and some are neither.
\[ f(x) = x^{3}, \qquad f'(x) = 3x^{2}, \qquad f'(0) = 0 \]
The origin is a critical point of the cubing function, and the tangent there really is horizontal. But the function is increasing on both sides of it.
\[ f'(x) = 3x^{2} > 0 \ \text{ for every } x \ne 0 \]
So the origin is neither a relative maximum nor a relative minimum. It is a flat spot on a rising curve.
Picture it
Animation
Shows: A critical point need not be an extremum — a rendered Manim animation.
Rendered with Manim.
Takeaway: The slope is zero at the origin, and the curve sails straight through.
Intuition
Think of a car that is speeding up, taps the brake for one instant until the speedometer reads exactly zero, and then immediately accelerates forward again.
The speed hit zero, but the car never reversed. Its position kept increasing the whole time, so that instant was not a maximum or a minimum of position.
A zero derivative tells you the motion paused. Only a change of direction - the derivative switching sign - makes it an extremum.
Fill the middle
Fill in the blanks
From Trap: every critical point is a max or a min — finish the line. Write what belongs on the right of the equals sign before you look.
h(x) = x^{3} - 3x^{2} + 3x
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The derivative is a perfect square trinomial, which factors to a squared binomial.
Trap
Classify the critical point of this function.
\[ h(x) = x^{3} - 3x^{2} + 3x \]
Differentiate and solve for the critical number
Why: The derivative is a perfect square trinomial, which factors to a squared binomial.
\[ h'(x) = 3x^{2} - 6x + 3 = 3(x-1)^{2}, \qquad x = 1 \]
Declare a relative minimum at the input 1
Why: Reasoning that a horizontal tangent must be a turning point. No sign test was performed, so the conclusion is a guess dressed up as a result.
\[ \text{``rel. min at } x=1\text{''} \quad \text{(wrong)} \]
Classify the critical point of this function.
\[ h(x) = x^{3} - 3x^{2} + 3x \]
Differentiate and solve for the critical number
Why: Same first move: the derivative factors as three times a squared binomial, giving one critical number.
\[ h'(x) = 3(x-1)^{2}, \qquad x = 1 \]
Test the sign of the derivative on each side
Why: A square is never negative, so the derivative is positive on both sides of 1. The function is increasing before and after.
| test input | value of the derivative | sign | behaviour |
|---|---|---|---|
| 0 | 3 | positive | increasing |
| 2 | 3 | positive | increasing |
No sign change, so it is neither a maximum nor a minimum
Why: The curve flattens for one instant and keeps climbing. This is the cubic's inflection point sitting on a horizontal tangent.
\[ x = 1 \ \text{is critical but not an extremum} \]
Comparison
Comparison matrix
From Trap: every critical point is a max or a min: refill the value of the derivative column from what you know. The rest of the table is as it appeared.
| test input | value of the derivative | sign | behaviour |
|---|---|---|---|
| 0 | 3 | positive | increasing |
| 2 | 3 | positive | increasing |
Section
Section 3
Pattern
The EVT promises the absolute extrema exist. Fermat says the interior ones hide at critical points. Endpoints cover the rest. That is a complete search.
No sign charts, no derivative tests. On a closed interval you only have to compare a short list of numbers.
Picture it
Animation
Shows: The candidate list, in full — a rendered Manim animation.
Rendered with Manim.
Takeaway: A finite list, which is why the method always terminates.
Estimation
Predict first
Find the absolute maximum and minimum values.
Commit before you compute: what does Worked example: closed-interval method on a cubic come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with a nearby test value
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Just to the right of the reported maximum, at the input negative 0.9, the function gives about 7.91, which is below 8; just to the left of the reported minimum, at 1.9, it gives about negative 18.91, which is above negative 19. Both extreme values behave the way a maximum and a minimum should.
Worked example
Find the absolute maximum and minimum values.
\[ f(x) = 2x^{3} - 3x^{2} - 12x + 1 \quad \text{on} \quad [-2, 3] \]
Check the hypotheses
Why: A polynomial is continuous everywhere, and the interval is closed and bounded, so the EVT guarantees both extrema exist and the method is legal.
Differentiate and factor
Why: Factoring the derivative turns the critical-point equation into two easy roots.
\[ f'(x) = 6x^{2} - 6x - 12 = 6(x-2)(x+1) \]
Solve for the critical numbers and keep the ones inside the interval
Why: Both roots lie strictly between negative 2 and 3, so neither gets discarded. The derivative is a polynomial, so there is no undefined case.
\[ x = -1 \ \text{ and } \ x = 2, \qquad \text{both in } (-2,3) \]
Evaluate f at the two critical numbers and the two endpoints
Why: Four inputs, four outputs. The absolute extrema must be among them - that is exactly what the EVT plus Fermat guarantee.
| input | computation | f value | role |
|---|---|---|---|
| -2 | -16 - 12 + 24 + 1 | -3 | endpoint |
| -1 | -2 - 3 + 12 + 1 | 8 | critical |
| 2 | 16 - 12 - 24 + 1 | -19 | critical |
| 3 | 54 - 27 - 36 + 1 | -8 | endpoint |
Compare the four outputs
Why: Largest of the list is 8; smallest is negative 19. Report both the value and where it happens.
\[ \text{abs. max } f(-1) = 8, \qquad \text{abs. min } f(2) = -19 \]
Verify with a nearby test value
Why: Just to the right of the reported maximum, at the input negative 0.9, the function gives about 7.91, which is below 8; just to the left of the reported minimum, at 1.9, it gives about negative 18.91, which is above negative 19. Both extreme values behave the way a maximum and a minimum should.
\[ f(-0.9) \approx 7.912 < 8, \qquad f(1.9) \approx -18.912 > -19 \]
Picture it
Animation
Shows: Each line of the worked example "closed-interval method on a cubic", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A polynomial is continuous everywhere, and the interval is closed and bounded, so the EVT guarantees both extrema exist and the method is legal.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Find the only critical number
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The derivative is a linear function, zero at the input 2, which is inside the interval.
Find the absolute extrema.
Why: The derivative is a linear function, zero at the input 2, which is inside the interval.
Trap
Find the absolute extrema.
\[ f(x) = x^{2} - 4x + 3 \quad \text{on} \quad [0, 3] \]
Find the only critical number
Why: The derivative is a linear function, zero at the input 2, which is inside the interval.
\[ f'(x) = 2x - 4 = 0 \ \Longrightarrow\ x = 2, \qquad f(2) = -1 \]
Report both extrema from that one point
Why: Treating the only candidate as the answer to both questions. But nothing was compared, because nothing else was evaluated.
\[ \text{``abs. max and abs. min both } = -1\text{''} \quad \text{(wrong)} \]
Find the absolute extrema.
\[ f(x) = x^{2} - 4x + 3 \quad \text{on} \quad [0, 3] \]
Find the critical number, then build the full candidate list
Why: The candidate list is always critical numbers PLUS both endpoints. Three inputs here, not one.
\[ \text{candidates: } x = 0, \ x = 2, \ x = 3 \]
Evaluate all three
Why: Only after every candidate has a number can you compare them.
| input | f value | role |
|---|---|---|
| 0 | 3 | endpoint |
| 2 | -1 | critical |
| 3 | 0 | endpoint |
Compare: the maximum lives at the left endpoint
Why: The parabola opens upward, so its vertex is the minimum and the maximum has to be pushed out to an endpoint. Missing the endpoints missed the entire maximum.
\[ \text{abs. max } f(0) = 3, \qquad \text{abs. min } f(2) = -1 \]
Pattern
Step through it
Step through Trap: forgetting the endpoints one row at a time. What is driving the change, and what would the row after the last one be?
Pattern
Predict first
The table runs: -1 | cube root of -1 is -1, squared | 1 | endpoint · 0 | cube root of 0 is 0, squared | 0 | critical
In Worked example: closed-interval method with a cusp, given the rows so far: what is the next one — the row where input is 8?
Correct: 8 | cube root of 8 is 2, squared | 4 | endpoint
| input | computation | f value | role |
|---|---|---|---|
| -1 | cube root of -1 is -1, squared | 1 | endpoint |
| 0 | cube root of 0 is 0, squared | 0 | critical |
| 8 | cube root of 8 is 2, squared | 4 | endpoint |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. The cube root is defined for every real number, including the negatives, so squaring it gives a function continuous on the whole interval.
Worked example
Find the absolute extrema.
\[ f(x) = x^{2/3} \quad \text{on} \quad [-1, 8] \]
Check continuity
Why: The cube root is defined for every real number, including the negatives, so squaring it gives a function continuous on the whole interval. The EVT applies.
Differentiate
Why: Power rule on the two-thirds exponent: subtract one to get negative one-third.
\[ f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \]
The derivative is never zero, but it is undefined at the origin
Why: The numerator is a nonzero constant, so case one gives nothing. The denominator vanishes at zero, and the function is defined there, so the origin is the only critical number.
\[ x = 0 \ \text{ is the only critical number, and } 0 \in (-1, 8) \]
Evaluate at the critical number and both endpoints
Why: Remember that a two-thirds power of a negative input is fine: cube root first, then square, so the output is positive.
| input | computation | f value | role |
|---|---|---|---|
| -1 | cube root of -1 is -1, squared | 1 | endpoint |
| 0 | cube root of 0 is 0, squared | 0 | critical |
| 8 | cube root of 8 is 2, squared | 4 | endpoint |
Compare
Why: Largest output is 4, smallest is 0. The minimum is exactly the cusp, which a zero-derivative-only search would have missed.
\[ \text{abs. max } f(8) = 4, \qquad \text{abs. min } f(0) = 0 \]
Verify the minimum by the shape of the formula
Why: A cube root squared can never be negative, and it equals zero only when the input is zero. So the value 0 really is the smallest possible output, everywhere - not just on this interval.
\[ x^{2/3} = \left(\sqrt[3]{x}\right)^{2} \ge 0 \ \text{ with equality only at } x = 0 \]
Picture it
Animation
Shows: Each line of the worked example "closed-interval method with a cusp", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The cube root is defined for every real number, including the negatives, so squaring it gives a function continuous on the whole interval. The EVT applies.
Elimination
Eliminate the wrong options
What is the absolute MINIMUM value of f on the closed interval from 0 to 3?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The derivative is three x squared minus three, which is zero at x equal to 1 and x equal to negative 1; only x equal to 1 lies in the interval. Evaluating gives f(0) = 0, f(1) = 1 - 3 = -2, and f(3) = 27 - 9 = 18. The smallest of 0, -2, and 18 is -2.
Check
Work it on paper with the closed-interval method before you choose.
\[ f(x) = x^{3} - 3x \quad \text{on} \quad [0, 3] \]
Check your understanding
What is the absolute MINIMUM value of f on the closed interval from 0 to 3?
Answer: A
Why: The derivative is three x squared minus three, which is zero at x equal to 1 and x equal to negative 1; only x equal to 1 lies in the interval. Evaluating gives f(0) = 0, f(1) = 1 - 3 = -2, and f(3) = 27 - 9 = 18. The smallest of 0, -2, and 18 is -2.
Check
Differentiate, factor out the most negative power, and check both cases.
\[ f(x) = x^{4/3} - 4x^{1/3} \]
Check your understanding
What are ALL of the critical numbers of f?
Answer: A
Why: The derivative is four-thirds times x to the one-third minus four-thirds times x to the negative two-thirds, which factors as four-thirds times x to the negative two-thirds times the quantity x minus 1. The numerator is zero at x equal to 1, and the derivative is undefined at x equal to 0 where f is still defined, so both are critical numbers.
Step zero
Discussion prompt
Worked example: absolute extrema of a product with a radical — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate with the product rule, chaining the radical
Answer:
Worked example
Find the absolute extrema on the natural domain of this function.
\[ f(x) = x\sqrt{4 - x^{2}} \quad \text{on} \quad [-2, 2] \]
Differentiate with the product rule, chaining the radical
Why: The first factor differentiates to 1; the radical differentiates to negative x over the radical by the chain rule.
\[ f'(x) = \sqrt{4-x^{2}} + x \cdot \frac{-x}{\sqrt{4-x^{2}}} \]
Put it over a common denominator
Why: Combining into one fraction makes both the zero case and the undefined case readable at a glance.
\[ f'(x) = \frac{(4-x^{2}) - x^{2}}{\sqrt{4-x^{2}}} = \frac{4 - 2x^{2}}{\sqrt{4-x^{2}}} \]
Set the numerator to zero
Why: Four minus two x squared is zero when x squared equals 2, giving the plus and minus square root of 2, both of which sit inside the interval.
\[ 4 - 2x^{2} = 0 \ \Longrightarrow\ x = \pm\sqrt{2} \approx \pm 1.414 \]
Note where the derivative is undefined
Why: The denominator vanishes at the inputs 2 and negative 2 - but those are the endpoints, not interior points, so they enter the candidate list as endpoints rather than as critical numbers.
Evaluate at all four candidates
Why: At the square root of 2, the radical also equals the square root of 2, so the product is exactly 2. The endpoints make the radical zero.
| input | computation | f value | role |
|---|---|---|---|
| -2 | -2 times 0 | 0 | endpoint |
| negative root 2 | negative root 2 times root 2 | -2 | critical |
| root 2 | root 2 times root 2 | 2 | critical |
| 2 | 2 times 0 | 0 | endpoint |
Compare the four values
Why: The list is 0, negative 2, 2, 0. Largest and smallest are immediate.
\[ \text{abs. max } f(\sqrt{2}) = 2, \qquad \text{abs. min } f(-\sqrt{2}) = -2 \]
Verify with a nearby input
Why: At the input 1.4, just left of the square root of 2, the function gives about 1.9996, which is below 2. The reported maximum really is a peak, and the odd symmetry of the formula makes the minimum its mirror image.
\[ f(1.4) = 1.4\sqrt{2.04} \approx 1.9996 < 2, \qquad f(-x) = -f(x) \]
Picture it
Animation
Shows: Absolute versus relative extrema — a rendered Manim animation.
Rendered with Manim.
Takeaway: Two local minima here, and only one of them is the best on the whole interval.
Section
Section 4
Intuition
Throw a ball straight up. It leaves your hand at shoulder height and lands back at shoulder height.
Its height went up and came back down, so somewhere in the middle its vertical velocity had to be exactly zero - at the top of the arc.
That is Rolle's Theorem in one sentence: same height at both ends forces a horizontal tangent somewhere in between.
Picture it
Figure (svg): A curve starting and ending at the same height with a dashed horizontal secant and a dashed horizontal tangent at the peak.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Three hypotheses go in, one guaranteed point comes out.
Concept
Three hypotheses go in, one guaranteed point comes out.
Figure (svg): A curve starting and ending at the same height with a dashed horizontal secant and a dashed horizontal tangent at the peak.
Rolle's Theorem — If f is continuous on the closed interval from a to b, differentiable on the open interval from a to b, and f(a) equals f(b), then there is at least one number c strictly between a and b with derivative zero at c.
\[ f \text{ cont. on } [a,b], \ \ f \text{ diff. on } (a,b), \ \ f(a)=f(b) \;\Longrightarrow\; \exists\, c \in (a,b): f'(c) = 0 \]
Notice the asymmetry: continuity is required on the closed interval, differentiability only on the open one. That lets vertical tangents at the endpoints slide through.
Missing information
Discussion prompt
Verify the hypotheses of Rolle's Theorem and find every value the theorem guarantees.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
A polynomial is continuous and differentiable at every real number, so it certainly is on the closed interval and on the open one.
Worked example
Verify the hypotheses of Rolle's Theorem and find every value the theorem guarantees.
\[ f(x) = x^{3} - x \quad \text{on} \quad [-1, 1] \]
Hypothesis 1 and 2: continuity and differentiability
Why: A polynomial is continuous and differentiable at every real number, so it certainly is on the closed interval and on the open one.
Hypothesis 3: check the two endpoint heights match
Why: Both endpoints output zero, so the equal-height condition holds and Rolle applies.
\[ f(-1) = -1 + 1 = 0, \qquad f(1) = 1 - 1 = 0 \]
Differentiate and set the derivative equal to zero
Why: The theorem promises a flat tangent, so the equation to solve is derivative equals zero.
\[ f'(x) = 3x^{2} - 1 = 0 \ \Longrightarrow\ x^{2} = \frac{1}{3} \]
Solve and keep the roots inside the open interval
Why: Both roots have absolute value about 0.577, comfortably strictly between negative 1 and 1, so both are legitimate. Rolle promised at least one; here there are two.
\[ c = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3} \approx \pm 0.577 \]
Verify by substituting back into the derivative
Why: Squaring either value gives one-third, so three times one-third minus 1 is zero. The tangent really is horizontal at both places.
\[ f'\!\left(\tfrac{\sqrt3}{3}\right) = 3\cdot\tfrac13 - 1 = 0 \]
Concept
The hypotheses are not decoration. Here is a function that satisfies two of the three and produces no guaranteed point at all.
\[ f(x) = x^{2/3} \quad \text{on} \quad [-1, 1], \qquad f(-1) = f(1) = 1 \]
It is continuous on the closed interval and the endpoint heights match. But it is not differentiable at the origin, which lies inside the open interval.
\[ f'(x) = \frac{2}{3\sqrt[3]{x}} \ne 0 \ \text{ for every } x, \quad f'(0) \ \text{undefined} \]
There is no horizontal tangent anywhere. The cusp turned the function around without ever going flat.
Picture it
Animation
Shows: Drop differentiability and Rolle fails — a rendered Manim animation.
Rendered with Manim.
Takeaway: The corner is exactly where the theorem's promise would have lived.
Concept
Rolle's Theorem is an existence statement. It says at least one such point exists; it never says exactly one.
The cubic example produced two. A wavy function with equal endpoint heights can easily produce a dozen.
\[ f(x) = \sin x \ \text{ on } \ [0, 4\pi]: \quad f(0) = f(4\pi) = 0 \]
\[ f'(x) = \cos x = 0 \ \text{ at } \ x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \tfrac{7\pi}{2} \]
Four guaranteed points on that interval. So when a problem says find the value the theorem guarantees, find them all and keep every one that lies strictly inside.
Section
Section 5
Intuition
You drive 120 miles in an hour and a half. Your average speed was 80 miles per hour.
You did not drive at a constant 80 the whole way - you stopped at a light, you crawled through a construction zone, you passed a truck. So sometimes you were slower than 80 and sometimes faster.
Since your speed moved continuously from below 80 to above 80, there had to be at least one instant when the speedometer read exactly 80.
That is the Mean Value Theorem. The average rate of change over the trip is achieved as an instantaneous rate of change at some moment during it.
Picture it
Figure (svg): A curve from a to b with a dashed secant line joining the endpoints and a dashed tangent line parallel to it at an interior point.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Take Rolle's Theorem and tilt it. The endpoints no longer have to match; the guaranteed slope is now the slope of the secant line.
Concept
Take Rolle's Theorem and tilt it. The endpoints no longer have to match; the guaranteed slope is now the slope of the secant line.
Figure (svg): A curve from a to b with a dashed secant line joining the endpoints and a dashed tangent line parallel to it at an interior point.
Mean Value Theorem (MVT) — If f is continuous on the closed interval from a to b and differentiable on the open interval from a to b, then there is at least one c strictly between a and b at which the instantaneous rate of change equals the average rate of change over the whole interval.
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
The right-hand side is a number you compute from just two function values. The left-hand side is a slope at one unknown place. The theorem says the two can be made equal.
Picture it
Animation
Shows: A secant across an interval matched by a parallel tangent inside it.
The tangent parallel to the secant.
Takeaway: Somewhere strictly inside the interval, the instantaneous slope equals the average slope across it. The theorem promises existence, not a location.
Concept
The fraction on the right is exactly the slope of the line through the two endpoints of the graph - the secant line.
\[ m_{\text{sec}} = \frac{f(b)-f(a)}{b-a} \]
So the MVT says: slide the secant line up or down, keeping its direction fixed, until it just touches the curve. Where it touches, it is tangent.
That touching point is the guaranteed value. Tangent parallel to secant is the picture to memorize.
Concept
The two theorems are not separate facts. Rolle is what the MVT says when the two endpoint heights happen to be equal.
\[ f(a) = f(b) \ \Longrightarrow\ \frac{f(b)-f(a)}{b-a} = \frac{0}{b-a} = 0 \]
The guaranteed slope collapses to zero, and the MVT's conclusion becomes exactly Rolle's conclusion.
\[ f'(c) = 0 \]
In the other direction, Rolle is the tool used to prove the MVT: subtract the secant line from the function and apply Rolle to the difference.
Picture it
Animation
Shows: What the MVT is actually for — a rendered Manim animation.
Rendered with Manim.
Takeaway: Almost every later theorem quietly leans on this one.
Picture it
Animation
Shows: The MVT and Rolle's theorem written one above the other.
One theorem, seen twice.
Takeaway: Rolle is what the Mean Value Theorem says when the endpoints happen to have equal heights. It is not a separate result to memorise.
Constraint
Discussion prompt
Run How to find the value the MVT guarantees with this step confiscated:
Set the derivative equal to that number and solve for the input.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Steps 2 and 4 are where students slip: the average rate uses function values, the instantaneous rate uses the derivative formula. Do not mix them.
Edge cases
Discussion prompt
How to find the value the MVT guarantees works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Steps 2 and 4 are where students slip: the average rate uses function values, the instantaneous rate uses the derivative formula. Do not mix them.
Ranking
Put in order
Put the moves of Worked example: the MVT on a cubic into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. A polynomial is continuous and differentiable everywhere, so both hypotheses hold automatically on any interval.
Worked example
Find all values guaranteed by the Mean Value Theorem.
\[ f(x) = x^{3} - 2x \quad \text{on} \quad [-1, 2] \]
Check the hypotheses
Why: A polynomial is continuous and differentiable everywhere, so both hypotheses hold automatically on any interval.
Evaluate at the two endpoints
Why: The average rate of change is built from function values, not derivatives, so these two numbers come first.
\[ f(-1) = -1 + 2 = 1, \qquad f(2) = 8 - 4 = 4 \]
Compute the average rate of change
Why: Rise over run across the whole interval: the change in output divided by the change in input.
\[ \frac{f(2)-f(-1)}{2-(-1)} = \frac{4-1}{3} = 1 \]
Differentiate and set the derivative equal to 1
Why: The theorem guarantees an input where the instantaneous slope matches that average slope of 1.
\[ f'(x) = 3x^{2} - 2 = 1 \ \Longrightarrow\ 3x^{2} = 3 \ \Longrightarrow\ x^{2} = 1 \]
Solve, then discard the root that is not strictly inside
Why: Both 1 and negative 1 solve the equation, but negative 1 is the left endpoint. The MVT only claims a point in the OPEN interval, so only 1 survives.
\[ x = \pm 1, \quad -1 \notin (-1,2) \ \Longrightarrow \ c = 1 \]
Verify the tangent slope matches the secant slope
Why: The derivative at 1 is three minus two, which is 1, exactly the average rate of change computed earlier. The tangent at that point is parallel to the secant through the endpoints.
\[ f'(1) = 3(1)^{2} - 2 = 1 \;=\; \frac{f(2)-f(-1)}{2-(-1)} \]
Picture it
Animation
Shows: Each line of the worked example "the MVT on a cubic", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A polynomial is continuous and differentiable everywhere, so both hypotheses hold automatically on any interval.
Hypothesis
Predict first
Worked example: the MVT with a vertical tangent at an endpoint is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Check the hypotheses carefully
Why: The square root is continuous on the closed interval including the left endpoint. Its derivative blows up at zero - but zero is an endpoint, and differentiability is only required on the OPEN interval. The hypotheses hold.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Find the value guaranteed by the Mean Value Theorem.
\[ f(x) = \sqrt{x} \quad \text{on} \quad [0, 4] \]
Check the hypotheses carefully
Why: The square root is continuous on the closed interval including the left endpoint. Its derivative blows up at zero - but zero is an endpoint, and differentiability is only required on the OPEN interval. The hypotheses hold.
\[ f \text{ cont. on } [0,4], \qquad f \text{ diff. on } (0,4) \]
Compute the average rate of change
Why: The square root of 4 is 2 and the square root of 0 is 0, so the rise is 2 over a run of 4.
\[ \frac{f(4)-f(0)}{4-0} = \frac{2-0}{4} = \frac{1}{2} \]
Differentiate
Why: Write the radical as a one-half power, then apply the power rule.
\[ f'(x) = \frac{1}{2\sqrt{x}} \]
Set the derivative equal to one-half and solve
Why: Cross-multiplying gives two times the square root of x equals 2, so the square root of x is 1, so x is 1.
\[ \frac{1}{2\sqrt{x}} = \frac{1}{2} \ \Longrightarrow\ \sqrt{x} = 1 \ \Longrightarrow\ c = 1 \]
Verify: the value lies inside and the slopes agree
Why: The number 1 is strictly between 0 and 4, and the derivative there is one over two times the square root of 1, which is one-half - exactly the average rate of change. The answer checks.
\[ f'(1) = \frac{1}{2\sqrt{1}} = \frac{1}{2}, \qquad 1 \in (0,4) \]
Picture it
Animation
Shows: Each line of the worked example "the MVT with a vertical tangent at an endpoint", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The square root is continuous on the closed interval including the left endpoint. Its derivative blows up at zero - but zero is an endpoint, and differentiability is only required on the OPEN interval. The hypotheses hold.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Find the value guaranteed by the Mean Value Theorem.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Skipping the hypothesis check because the formula looks harmless.
Find the value guaranteed by the Mean Value Theorem.
Why: Skipping the hypothesis check because the formula looks harmless. The endpoint outputs are 1 and negative 1.
Trap
Find the value guaranteed by the Mean Value Theorem.
\[ f(x) = \frac{1}{x} \quad \text{on} \quad [-1, 1] \]
Jump straight to the average rate of change
Why: Skipping the hypothesis check because the formula looks harmless. The endpoint outputs are 1 and negative 1.
\[ \frac{f(1)-f(-1)}{1-(-1)} = \frac{1-(-1)}{2} = 1 \]
Set the derivative equal to 1 and solve
Why: The derivative is negative one over x squared. Setting it equal to 1 gives x squared equal to negative 1, which has no real solution.
\[ -\frac{1}{x^{2}} = 1 \ \Longrightarrow\ x^{2} = -1 \ \ \text{(no real solution)} \]
Conclude the Mean Value Theorem is false
Why: The real problem is upstream: the theorem was never applicable, so its conclusion was never promised. The derivative is negative everywhere it exists, so it can never equal a positive 1.
Find the value guaranteed by the Mean Value Theorem.
\[ f(x) = \frac{1}{x} \quad \text{on} \quad [-1, 1] \]
Check continuity on the closed interval FIRST
Why: The reciprocal function is undefined at zero, and zero sits inside this interval. So f is not even defined on all of the closed interval, let alone continuous on it.
\[ 0 \in [-1,1], \quad f(0) \ \text{undefined}, \quad \lim_{x \to 0^{+}} \frac{1}{x} = \infty \]
Stop: the MVT does not apply, so it guarantees nothing
Why: A theorem whose hypotheses fail makes no claim at all. Finding no such value is not a contradiction; it is exactly what an inapplicable theorem allows.
Restrict to an interval where the hypotheses hold
Why: On the interval from 1 to 3 the function is continuous and differentiable. The average rate is negative one-third, and solving gives the square root of 3, which lies strictly inside. The theorem works fine when its hypotheses do.
\[ \text{on } [1,3]: \ \frac{\frac13 - 1}{2} = -\frac13, \quad -\frac{1}{x^{2}} = -\frac13 \ \Rightarrow\ c = \sqrt{3} \approx 1.73 \]
Notation
Annotate
From Trap: using the MVT across a discontinuity — read this one piece at a time. What is each part doing?
On: \( \text{on } [1,3]: \ \frac{\frac13 - 1}{2} = -\frac13, \quad -\frac{1}{x^{2}} = -\frac13 \ \Rightarrow\ c = \sqrt{3} \approx 1.73 \)
Prediction
Predict first
Which value of c satisfies the conclusion of the Mean Value Theorem on this interval?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: c = 9/4
Why: The average rate of change is the square root of 9 minus the square root of 0, all over 9, which is one-third. Setting the derivative one over two root x equal to one-third gives two root x equal to 3, so root x is three-halves and c is nine-fourths, which lies strictly between 0 and 9.
Check
Compute the average rate of change, then set the derivative equal to it.
\[ f(x) = \sqrt{x} \quad \text{on} \quad [0, 9] \]
Check your understanding
Which value of c satisfies the conclusion of the Mean Value Theorem on this interval?
Answer: A
Why: The average rate of change is the square root of 9 minus the square root of 0, all over 9, which is one-third. Setting the derivative one over two root x equal to one-third gives two root x equal to 3, so root x is three-halves and c is nine-fourths, which lies strictly between 0 and 9.
Step zero
Discussion prompt
Worked example: the MVT writes a speeding ticket — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the function and the interval, measuring time in hours
Answer:
Worked example
A car enters a toll road at 1:00 pm and exits 120 miles later at 2:30 pm. The posted speed limit is 65 miles per hour. Prove the car was speeding at some instant.
Name the function and the interval, measuring time in hours
Why: Position as a function of time is continuous and differentiable - a car cannot teleport or change speed instantaneously - so the MVT hypotheses hold.
\[ s(0) = 0 \ \text{miles}, \qquad s(1.5) = 120 \ \text{miles} \]
Compute the average rate of change
Why: Change in position over change in time is exactly average velocity, in miles per hour.
\[ \frac{s(1.5)-s(0)}{1.5-0} = \frac{120}{1.5} = 80 \ \text{mi/hr} \]
Apply the MVT to convert the average into an instant
Why: The theorem guarantees a time strictly inside the trip at which the derivative of position - the instantaneous speed on the speedometer - equals that average.
\[ \exists\, c \in (0, 1.5): \quad s'(c) = 80 \ \text{mi/hr} \]
Compare against the limit
Why: Eighty is greater than sixty-five, so at that instant the car exceeded the posted limit. The toll booth timestamps alone are a proof.
\[ 80 > 65 \ \Longrightarrow\ \text{speeding at time } c \]
Verify the units and the logic
Why: Miles divided by hours gives miles per hour, matching a speed limit's units. And the argument does not need to know the speed at any other moment - only that the trip's average was above the limit, which forces at least one instant at that exact speed.
Picture it
Animation
Shows: The MVT writes speeding tickets — a rendered Manim animation.
Rendered with Manim.
Takeaway: Average speed forces an instant that matches it.
Section
Section 6
Concept
It feels obvious that a function with no slope anywhere cannot move. The MVT is what makes that obvious feeling a theorem.
\[ f'(x) = 0 \ \text{ for all } x \in (a,b) \;\Longrightarrow\; f \text{ is constant on } (a,b) \]
Without the MVT there is no bridge from a statement about slopes at every point to a statement about values at every point. This is the bridge.
Estimation
Predict first
Show that a function whose derivative is zero throughout an interval must take the same value at every point of that interval.
Commit before you compute: what does Worked example: proving the zero-derivative fact come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the theorem on a concrete case
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The sum of sine squared and cosine squared has derivative two sine cosine minus two cosine sine, which is zero everywhere.
Worked example
Show that a function whose derivative is zero throughout an interval must take the same value at every point of that interval.
Pick any two points in the interval and call the smaller one first
Why: To prove the function is constant it is enough to show any two of its values are equal. Naming them lets the MVT act on the closed interval between them.
\[ x_{1} < x_{2}, \qquad [x_{1}, x_{2}] \subseteq I \]
Apply the MVT on that sub-interval
Why: The function is differentiable on the whole interval, so it is continuous there too; both hypotheses hold on the smaller interval as well.
\[ \exists\, c \in (x_{1}, x_{2}): \quad f'(c) = \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}} \]
Substitute the hypothesis that the derivative is zero
Why: The derivative is zero at every point, and c is a point, so the left side of the equation is zero.
\[ 0 = \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}} \]
Multiply through by the nonzero denominator
Why: The two inputs are different, so their difference is not zero and multiplying is legal. The numerator must therefore be zero.
\[ f(x_{2}) - f(x_{1}) = 0 \ \Longrightarrow\ f(x_{2}) = f(x_{1}) \]
Note the two points were arbitrary
Why: Nothing in the argument depended on which two points were chosen, so all values agree and the function is constant.
Verify the theorem on a concrete case
Why: The sum of sine squared and cosine squared has derivative two sine cosine minus two cosine sine, which is zero everywhere. The theorem therefore predicts it is constant - and indeed it is always 1, the Pythagorean identity.
\[ \frac{d}{dx}\left[\sin^{2}x + \cos^{2}x\right] = 2\sin x\cos x - 2\cos x\sin x = 0 \ \Longrightarrow\ \sin^{2}x + \cos^{2}x = 1 \]
Picture it
Animation
Shows: Each line of the worked example "proving the zero-derivative fact", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum of sine squared and cosine squared has derivative two sine cosine minus two cosine sine, which is zero everywhere. The theorem therefore predicts it is constant - and indeed it is always 1, the Pythagorean identity.
Concept
Apply the previous fact to a difference and a second, larger consequence falls out immediately.
\[ f'(x) = g'(x) \ \text{ on } I \;\Longrightarrow\; f(x) = g(x) + C \ \text{ on } I \]
The proof is one line: the difference of the two functions has derivative zero, so by the previous result the difference is a constant.
\[ (f-g)'(x) = f'(x) - g'(x) = 0 \ \Longrightarrow\ f - g = C \]
Concept
This is the theorem that justifies the arbitrary constant you will write on every indefinite integral in the next chapter.
Once you find one function whose derivative is the one you want, this result says you have found all of them - they differ from yours by a constant and nothing more.
\[ \int 2x \, dx = x^{2} + C \]
Without the MVT you could not rule out some exotic extra antiderivative hiding somewhere. With it, the family is completely described.
Explain it
Discussion prompt
Explain Why that constant matters later to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Once you find one function whose derivative is the one you want, this result says you have found all of them - they differ from yours by a constant and nothing more.
Section
Section 7
Intuition
Walk left to right along the graph. If the tangent line tilts uphill, you are climbing. If it tilts downhill, you are descending.
The derivative is that tilt as a number. Its sign - positive or negative - is all you need to know which way the function is going.
So the whole shape story of a function is encoded in one question asked repeatedly: is the derivative above or below zero here?
Analogy
Discussion prompt
Explain The sign of the slope is the direction of travel by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Walk left to right along the graph. If the tangent line tilts uphill, you are climbing. If it tilts downhill, you are descending.
Concept
This too is a corollary of the Mean Value Theorem, proved the same way as the constant result.
\[ f'(x) > 0 \ \text{ on } (a,b) \;\Longrightarrow\; f \text{ is increasing on } (a,b) \]
\[ f'(x) < 0 \ \text{ on } (a,b) \;\Longrightarrow\; f \text{ is decreasing on } (a,b) \]
The proof: take any two inputs, apply the MVT, and note that a positive derivative times a positive run forces the outputs to rise.
Counterexample
Discussion prompt
The proof: take any two inputs, apply the MVT, and note that a positive derivative times a positive run forces the outputs to rise.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Pattern
Use the factored form when you test. Counting negative factors is faster and far less error-prone than evaluating an expanded polynomial.
Estimation
Predict first
Find the intervals on which this function is increasing and decreasing.
Commit before you compute: what does Worked example: where is the function increasing? come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with actual function values
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Moving from the input negative 3 to negative 2 the output rises from negative 22 to 3, matching increasing.
Worked example
Find the intervals on which this function is increasing and decreasing.
\[ f(x) = x^{3} - 3x^{2} - 9x + 5 \]
Differentiate and factor
Why: This is the cubic from earlier; its derivative factors into three times two linear factors.
\[ f'(x) = 3x^{2} - 6x - 9 = 3(x-3)(x+1) \]
Mark the split points on a number line
Why: The derivative is a polynomial, so it is never undefined. The only sign-flip candidates are its two roots.
\[ x = -1 \quad \text{and} \quad x = 3 \]
Test one value in each of the three intervals
Why: Substitute into the FACTORED derivative and track the sign of each factor, not the arithmetic.
| interval | test input | sign of (x-3) | sign of (x+1) | sign of the derivative | behaviour |
|---|---|---|---|---|---|
| left of -1 | -2 | negative | negative | positive | increasing |
| between -1 and 3 | 0 | negative | positive | negative | decreasing |
| right of 3 | 4 | positive | positive | positive | increasing |
State the answer as intervals
Why: Report open intervals of inputs, not outputs. The two critical numbers are the boundaries.
\[ \text{increasing on } (-\infty,-1) \cup (3,\infty), \qquad \text{decreasing on } (-1,3) \]
Verify with actual function values
Why: Moving from the input negative 3 to negative 2 the output rises from negative 22 to 3, matching increasing. Moving from 0 to 1 the output falls from 5 to negative 6, matching decreasing. The chart agrees with the function.
\[ f(-3) = -22 < f(-2) = 3; \qquad f(0) = 5 > f(1) = -6 \]
Picture it
Animation
Shows: Each line of the worked example "where is the function increasing?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Moving from the input negative 3 to negative 2 the output rises from negative 22 to 3, matching increasing. Moving from 0 to 1 the output falls from 5 to negative 6, matching decreasing. The chart agrees with the function.
Concept
Once the sign chart is built, classifying a critical point takes no extra work - just read the two signs beside it.
First Derivative Test — At a critical number c: if the derivative changes from positive to negative there, f has a relative maximum at c; if it changes from negative to positive, a relative minimum; if it does not change sign, c is neither.
| sign of the derivative before c | sign after c | conclusion at c |
|---|---|---|
| positive | negative | relative maximum |
| negative | positive | relative minimum |
| positive | positive | neither - still increasing |
| negative | negative | neither - still decreasing |
This test works at every critical number, including the ones where the derivative is undefined. That is its advantage over the second-derivative test you will meet next deck.
Comparison
Comparison matrix
From The First Derivative Test: refill the sign after c column from what you know. The rest of the table is as it appeared.
| sign of the derivative before c | sign after c | conclusion at c |
|---|---|---|
| positive | negative | relative maximum |
| negative | positive | relative minimum |
| positive | positive | neither - still increasing |
| negative | negative | neither - still decreasing |
Step zero
Discussion prompt
Worked example: classifying both critical points — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Read the sign chart at the first critical number
Answer:
Worked example
Classify each critical number of the same cubic and give the relative extreme values.
\[ f(x) = x^{3} - 3x^{2} - 9x + 5, \qquad f'(x) = 3(x-3)(x+1) \]
Read the sign chart at the first critical number
Why: The derivative is positive to the left of negative 1 and negative to the right of it. Positive then negative is the maximum pattern.
\[ f' : \ + \ \longrightarrow \ - \ \text{ at } x = -1 \]
Compute the relative maximum VALUE
Why: The test locates the extremum; you still have to evaluate the original function to report the value.
\[ f(-1) = -1 - 3 + 9 + 5 = 10 \]
Read the sign chart at the second critical number
Why: The derivative is negative to the left of 3 and positive to the right. Negative then positive is the minimum pattern.
\[ f' : \ - \ \longrightarrow \ + \ \text{ at } x = 3 \]
Compute the relative minimum VALUE
Why: Evaluate the original cubic at 3: twenty-seven minus twenty-seven minus twenty-seven plus five.
\[ f(3) = 27 - 27 - 27 + 5 = -22 \]
Verify each classification against neighbouring outputs
Why: Near the reported maximum, the outputs at negative 2 and at 0 are 3 and 5, both below 10. Near the reported minimum, the outputs at 2 and at 4 are negative 17 and negative 15, both above negative 22. Both classifications hold.
| input | f value | compared to the extremum |
|---|---|---|
| -2 | 3 | below the max of 10 |
| 0 | 5 | below the max of 10 |
| 2 | -17 | above the min of -22 |
| 4 | -15 | above the min of -22 |
Picture it
Animation
Shows: Each line of the worked example "classifying both critical points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Near the reported maximum, the outputs at negative 2 and at 0 are 3 and 5, both below 10. Near the reported minimum, the outputs at 2 and at 4 are negative 17 and negative 15, both above negative 22. Both classifications hold.
Ranking
Put in order
Put the moves of Worked example: a relative maximum at a cusp into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The numerator vanishes at the input 2; the denominator vanishes at the origin, where the function is still defined.
Worked example
Classify every critical number of this function.
\[ f(x) = x^{5/3} - 5x^{2/3}, \qquad f'(x) = \frac{5(x-2)}{3\sqrt[3]{x}} \]
Recall the two critical numbers
Why: The numerator vanishes at the input 2; the denominator vanishes at the origin, where the function is still defined.
\[ x = 0 \ \text{(derivative undefined)}, \qquad x = 2 \ \text{(derivative zero)} \]
Test the sign of the derivative in all three intervals
Why: Below zero, the cube root is negative and so is the numerator, so the quotient is positive. Between zero and 2 the cube root is positive but the numerator is still negative. Above 2 both are positive.
| interval | test input | value of the derivative | sign | behaviour |
|---|---|---|---|---|
| left of 0 | -1 | 5 | positive | increasing |
| between 0 and 2 | 1 | -5/3 | negative | decreasing |
| right of 2 | 8 | 5 | positive | increasing |
Apply the First Derivative Test at the origin
Why: Positive then negative is the maximum pattern - and this happens at a point where there is no tangent line at all. A test that only looked for zero derivatives would have missed it.
\[ \text{rel. max at } x=0, \qquad f(0) = 0 \]
Apply the test at the input 2
Why: Negative then positive is the minimum pattern. Factor the two-thirds power out of the value to keep it exact.
\[ f(2) = 2^{2/3}(2 - 5) = -3\sqrt[3]{4} \approx -4.76 \]
Verify the extreme values numerically
Why: At the input 1 the function gives 1 minus 5, which is negative 4, below the reported maximum of 0 and above the reported minimum of about negative 4.76. Both classifications are consistent with a real function value between them.
\[ f(1) = 1 - 5 = -4, \qquad -4.76 < -4 < 0 \]
Picture it
Animation
Shows: Each line of the worked example "a relative maximum at a cusp", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the input 1 the function gives 1 minus 5, which is negative 4, below the reported maximum of 0 and above the reported minimum of about negative 4.76. Both classifications are consistent with a real function value between them.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Classify the critical number of this function.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Substituting 1 and 3 into the ORIGINAL function gives negative 3 both times - the wrong function was sampled.
Classify the critical number of this function.
Why: Substituting 1 and 3 into the ORIGINAL function gives negative 3 both times - the wrong function was sampled.
Trap
Classify the critical number of this function.
\[ f(x) = x^{2} - 4x, \qquad f'(x) = 2x - 4, \qquad x = 2 \]
Plug test inputs into f and compare signs
Why: Substituting 1 and 3 into the ORIGINAL function gives negative 3 both times - the wrong function was sampled.
| test input | value of f | sign |
|---|---|---|
| 1 | -3 | negative |
| 3 | -3 | negative |
Conclude there is no sign change, so no extremum
Why: The signs of f itself say nothing about whether f is rising or falling. This test answered a question nobody asked.
\[ \text{``neither a max nor a min''} \quad \text{(wrong)} \]
Classify the critical number of this function.
\[ f(x) = x^{2} - 4x, \qquad f'(x) = 2x - 4, \qquad x = 2 \]
Plug the test inputs into the DERIVATIVE
Why: The First Derivative Test is about the slope on each side, so the derivative formula is the one to sample.
| test input | value of the derivative | sign | behaviour |
|---|---|---|---|
| 1 | -2 | negative | decreasing |
| 3 | 2 | positive | increasing |
Negative then positive: relative minimum
Why: The function falls into the input 2 and rises out of it, which is exactly the shape of a valley.
\[ \text{rel. min } f(2) = 4 - 8 = -4 \]
Sanity-check against the parabola's vertex
Why: Completing the square gives the vertex form, whose vertex sits at the input 2 with output negative 4. The derivative test and the algebra agree.
\[ f(x) = (x-2)^{2} - 4 \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Check
Factor the derivative and test one input on each side of the critical number.
\[ f(x) = x^{3} - 12x \]
Check your understanding
Classify the critical number at x = -2.
Answer: A
Why: The derivative is three x squared minus 12, which factors as three times the quantity x minus 2 times the quantity x plus 2. Testing the input -3 gives a positive derivative and testing 0 gives a negative one, so the sign goes positive to negative and x equal to -2 is a relative maximum, with f(-2) = -8 + 24 = 16.
Elimination
Eliminate the wrong options
Which statement about the Extreme Value Theorem is correct for this function on this interval?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The function is undefined at the input 2, and 2 lies inside the interval from 1 to 3, so f has an infinite discontinuity there. The EVT requires continuity on the entire closed interval, so its guarantee is void - and in fact the outputs are unbounded in both directions near 2.
Check
Before quoting any theorem, test its hypotheses against the function you were handed.
\[ f(x) = \frac{1}{x-2} \quad \text{on} \quad [1, 3] \]
Check your understanding
Which statement about the Extreme Value Theorem is correct for this function on this interval?
Answer: A
Why: The function is undefined at the input 2, and 2 lies inside the interval from 1 to 3, so f has an infinite discontinuity there. The EVT requires continuity on the entire closed interval, so its guarantee is void - and in fact the outputs are unbounded in both directions near 2.
Pattern
Four tools, four different jobs. Match the question to the tool before computing anything.
| The question | The tool | What you must check first |
|---|---|---|
| Do extrema even exist here? | Extreme Value Theorem | continuous on a closed, bounded interval |
| What are the absolute extrema on a closed interval? | closed-interval method | same as the EVT, then find ALL critical numbers |
| Is there a point with a horizontal tangent? | Rolle's Theorem | continuous on closed, differentiable on open, equal endpoint values |
| Is there a point whose slope equals the average slope? | Mean Value Theorem | continuous on closed, differentiable on open |
| Is this critical point a max, a min, or neither? | First Derivative Test | the sign of the derivative on both sides |
Every row's third column is where the marks are lost on an exam. Write the hypothesis check down; do not do it in your head.
Trade off
Comparison matrix
From Which theorem do I reach for?: every row here is a choice with a cost. Fill the The tool column, then say which row you would actually pick and what you give up for it.
| The question | The tool | What you must check first |
|---|---|---|
| Do extrema even exist here? | Extreme Value Theorem | continuous on a closed, bounded interval |
| What are the absolute extrema on a closed interval? | closed-interval method | same as the EVT, then find ALL critical numbers |
| Is there a point with a horizontal tangent? | Rolle's Theorem | continuous on closed, differentiable on open, equal endpoint values |
| Is there a point whose slope equals the average slope? | Mean Value Theorem | continuous on closed, differentiable on open |
| Is this critical point a max, a min, or neither? | First Derivative Test | the sign of the derivative on both sides |
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Part 1 - Absolute and Relative Extreme Values · Part 2 - Critical Points and Fermat's Theorem · Part 3 - The Closed-Interval Method · Part 4 - Rolle's Theorem · Part 5 - The Mean Value Theorem · Part 6 - What the MVT Buys You. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
| If you see | Do this |
|---|---|
| absolute extrema on a closed interval | critical numbers plus both endpoints, compare values |
| a fractional or negative exponent | check where the derivative is undefined too |
| equal function values at both ends | Rolle: solve derivative equals zero |
| find the c guaranteed by the theorem | set the derivative equal to the average rate of change |
| classify a critical point | sign of the derivative on each side |
Next deck: the second derivative - concavity, inflection points, the Second Derivative Test, and full curve sketching.
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