This deck explains what the second derivative measures and distinguishes concave up from concave down. It covers inflection points and the sign change they require, the Second Derivative Test and its inconclusive case, and first- and second-derivative sign charts, then works the full curve-sketching checklist end to end on a polynomial, a rational function, and an exponential. It targets the classic errors: calling every zero of the second derivative an inflection point, confusing decreasing with concave down, treating an inconclusive test as proof that there is no extremum, and reading features of a function off the graph of its derivative.
Subject: Calculus I · 124 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 15
The second derivative, inflection points, and drawing a graph you can defend
Objectives
By the end of this deck you will be able to:
Warm-up
Discussion prompt
Before we open Concavity, Inflection Points, and Curve Sketching: without looking back, what was the main idea of Extreme Values, Rolle's Theorem, and the MVT, 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 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.
Section
Part 1
Concept
The first derivative is a function. Like any function, it has its own derivative.
\[ f''(x) = \frac{d}{dx}\left[\, f'(x) \,\right] \]
So the second derivative answers one question: is the slope going up or down?
Counterexample
Discussion prompt
The first derivative is a function. Like any function, it has its own derivative.
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.
Intuition
Think of a car on a straight road. The position function is the odometer: how far you have gone.
The first derivative is the speedometer: how fast the odometer reading is changing.
The second derivative is the accelerator pedal: how fast the speedometer needle is changing. Press down and the needle climbs; that is a positive second derivative.
Notice you can be moving forward fast while the needle drops. Speed and change-in-speed are separate facts. Hold on to that - it is the single most useful idea in this deck.
Analogy
Discussion prompt
Explain Odometer, speedometer, accelerator 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:
Think of a car on a straight road. The position function is the odometer: how far you have gone.
Concept
A curve is concave up on an interval when its slopes get larger as you move left to right.
The slope can go from very negative, to mildly negative, to zero, to positive. Every one of those is an increase.
concave up — On an interval, the derivative is an increasing function there. The curve bends upward, like the inside of a bowl.
Explain it
Discussion prompt
Explain Concave up: the slopes are increasing 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:
A curve is concave up on an interval when its slopes get larger as you move left to right.
Picture it
Animation
Shows: Concave up: the slopes are increasing — a rendered Manim animation.
Rendered with Manim.
Takeaway: Steeply down, then flat, then steeply up — the slope only ever rises.
Picture it
Animation
Shows: The sign rules for the first and second derivatives side by side.
Rising and bending are independent.
Takeaway: The first derivative says rising or falling; the second says which way it bends. A function can rise while curving downward — the two are independent.
Concept
A curve is concave down on an interval when its slopes get smaller as you move left to right.
concave down — On an interval, the derivative is a decreasing function there. The curve bends downward, like the outside of a dome.
Definition probe
Sort into buckets
Every line below is part of the definition of concave up or of concave down — one or the other, never both. Put each where it belongs.
Intuition
Figure (svg): Two curves side by side: a U-shaped curve labelled concave up and an arch labelled concave down
The cheapest mental picture in all of calculus: a concave-up piece of curve holds water. A concave-down piece spills it.
Pour water on the left-hand shape and it pools. Pour it on the right-hand shape and it runs off both sides.
Use it to check yourself in three seconds on any exam sketch.
Picture it
Figure (svg): A U-shaped curve with a tangent line touching it at one point and lying entirely below the curve
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:
Draw the tangent line at any point of a concave-up arc. The curve stays above that line near the point.
Concept
Draw the tangent line at any point of a concave-up arc. The curve stays above that line near the point.
On a concave-down arc the curve stays below its tangent line.
Figure (svg): A U-shaped curve with a tangent line touching it at one point and lying entirely below the curve
This is exactly why a tangent-line estimate under-shoots a concave-up function and over-shoots a concave-down one.
Concept
Increasing slopes means the derivative is increasing, and a function is increasing where its derivative is positive. Apply that to the derivative itself.
\[ f''(x) > 0 \text{ on } I \;\Longrightarrow\; f \text{ is concave up on } I \]
\[ f''(x) < 0 \text{ on } I \;\Longrightarrow\; f \text{ is concave down on } I \]
That is the whole test. Concavity questions are sign-of-the-second-derivative questions.
Picture it
Animation
Shows: A cubic that is concave down then concave up.
The switch is the inflection point.
Takeaway: Concave down, then concave up. The point where the bend reverses is the inflection point, and the second derivative changes sign there.
Intuition
Walk along the curve from left to right and read the slope at each step, as if the slope were a number on a display.
If that display keeps ticking upward, each step tilts you a little more uphill than the last. Repeat that and the path curls upward. That is concave up.
If the display keeps ticking downward, each step tilts you a little more downhill than the last, and the path curls over. That is concave down.
Concept
The first derivative tells you the direction of travel. The second derivative tells you the direction of the bend. All four combinations happen.
| First derivative | Second derivative | What the graph does |
|---|---|---|
| positive | positive | rising, and rising faster and faster |
| positive | negative | rising, but levelling off |
| negative | positive | falling, but the fall is flattening out |
| negative | negative | falling, and falling ever more steeply |
Row three is the one students get wrong. Falling and holding water at the same time is completely normal.
Comparison
Comparison matrix
From Rising or falling is a different question from bending: refill the Second derivative column from what you know. The rest of the table is as it appeared.
| First derivative | Second derivative | What the graph does |
|---|---|---|
| positive | positive | rising, and rising faster and faster |
| positive | negative | rising, but levelling off |
| negative | positive | falling, but the fall is flattening out |
| negative | negative | falling, and falling ever more steeply |
Anomaly
Predict first
A student writes this, and it looks reasonable:
The reasoning: this graph goes down, so it must be concave down.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The word down is doing all the work here, and it is being used for two different things: direction of travel and direction of bend.
Compute both derivatives and read their signs separately.
Why: The word down is doing all the work here, and it is being used for two different things: direction of travel and direction of bend.
Trap
The reasoning: this graph goes down, so it must be concave down.
\[ f(x) = \frac{1}{x}, \qquad x > 0 \]
Declare the graph concave down because it falls
Why: The word down is doing all the work here, and it is being used for two different things: direction of travel and direction of bend.
\[ \text{claim: } f''(x) < 0 \quad \text{(false)} \]
Compute both derivatives and read their signs separately.
\[ f(x) = x^{-1}, \quad f'(x) = -x^{-2}, \quad f''(x) = 2x^{-3} \]
Read the first derivative: negative for every positive x
Why: Negative one over x squared is negative, so the graph really is falling on this interval.
Read the second derivative: positive for every positive x
Why: Two over x cubed is positive when x is positive, so the graph is concave up. Falling and concave up together.
| x | f(x) | drop from the previous value |
|---|---|---|
| 1 | 1 | - |
| 2 | 0.5 | 0.5 |
| 3 | 0.3333 | 0.1667 |
| 4 | 0.25 | 0.0833 |
The drops shrink every step. The graph is falling less and less steeply - which is exactly what increasing slopes look like.
Pattern
Step through it
Step through Trap: reading a falling graph as concave down one row at a time. What is driving the change, and what would the row after the last one be?
Pattern
The sign cannot change inside an interval, because you already collected every place where it could flip. That is why testing one point per interval is enough.
Picture it
Figure (svg): Number line split at x equals 1, marked negative and concave down on the left, positive and concave up on the right
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:
Find the intervals of concavity for this function.
Worked example
Find the intervals of concavity for this function.
\[ f(x) = x^3 - 3x^2 + 1 \]
Differentiate once
Why: Power rule term by term; the constant 1 contributes nothing.
\[ f'(x) = 3x^2 - 6x \]
Differentiate again and factor
Why: Factoring makes the sign obvious at a glance, which is the only thing a sign chart needs.
\[ f''(x) = 6x - 6 = 6(x-1) \]
Set the second derivative to zero to find the split point
Why: It is a polynomial, so it is defined everywhere; the only place the sign can flip is where it equals zero.
\[ 6(x-1) = 0 \;\Longrightarrow\; x = 1 \]
Test one point on each side
Why: Use x equals 0 on the left and x equals 2 on the right - both are easy to evaluate.
\[ f''(0) = -6 < 0, \qquad f''(2) = 6 > 0 \]
Figure (svg): Number line split at x equals 1, marked negative and concave down on the left, positive and concave up on the right
State the answer in interval form
Why: The split point itself belongs to neither open interval; concavity is a statement about intervals.
\[ \text{concave down on } (-\infty, 1), \qquad \text{concave up on } (1, \infty) \]
Verify with two extra test points
Why: Take x equals one half: six times one half minus six is negative three, still negative. Take x equals three halves: nine minus six is three, still positive. Both agree with the chart.
\[ f''\left(\tfrac{1}{2}\right) = -3 < 0, \qquad f''\left(\tfrac{3}{2}\right) = 3 > 0 \]
Notation
Annotate
From Worked example: concavity of a cubic — read this one piece at a time. What is each part doing?
On: \( f(x) = x^3 - 3x^2 + 1 \)
Elimination
Eliminate the wrong options
On the interval where x is positive, this function is:
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: Rewrite as x to the negative one. The first derivative is negative one over x squared, which is negative for every positive x, so the graph is decreasing. The second derivative is two over x cubed, which is positive for every positive x, so the graph is concave up.
Check
Work out both derivatives before you choose.
\[ f(x) = \frac{1}{x} \quad \text{on the interval } x > 0 \]
Check your understanding
On the interval where x is positive, this function is:
Answer: A
Why: Rewrite as x to the negative one. The first derivative is negative one over x squared, which is negative for every positive x, so the graph is decreasing. The second derivative is two over x cubed, which is positive for every positive x, so the graph is concave up.
Section
Part 2
Concept
An inflection point is a point on the graph where the concavity changes from up to down, or from down to up.
inflection point — A point on the curve at which the function is continuous and the concavity actually changes sides. Both halves matter: the point must be on the graph, and the bend must genuinely switch.
Because it is a point, report it as a coordinate pair, not just an x-value - you have to substitute back into the original function.
Intuition
Drive a road that curves left, then curves right. There is one instant where the wheel passes through straight-ahead on its way from one lock to the other.
That instant is the inflection point. You are still moving - possibly fast - but the curvature has just flipped sides.
Now the crucial part: if you turn the wheel toward straight and then turn back the same way, you never inflected. Touching zero is not enough. The wheel has to come out the other side.
Concept
Concavity can only flip where the second derivative is zero or where it fails to exist.
\[ f''(c) = 0 \qquad \text{or} \qquad f''(c) \text{ does not exist} \]
These are candidates only. The word is deliberate: a candidate has to win an election before it becomes an inflection point.
Concept
A candidate becomes an inflection point only if the second derivative has opposite signs on the two sides of it.
Same sign on both sides means the curve bent the same way going in and coming out. Nothing inflected.
There is one more requirement, and it is easy to forget: the point has to be on the graph. If the function is not even defined there, there is no point to name.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The reasoning: set the second derivative to zero, solve, and report the answers as inflection points.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The equation does give x equals 0.
Solve for the candidate, then test the sign on both sides before you claim anything.
Why: The equation does give x equals 0. The mistake is treating that as the finish line instead of the starting line.
Trap
The reasoning: set the second derivative to zero, solve, and report the answers as inflection points.
\[ f(x) = x^4, \qquad f''(x) = 12x^2 \]
Solve twelve x squared equals zero and stop
Why: The equation does give x equals 0. The mistake is treating that as the finish line instead of the starting line.
\[ 12x^2 = 0 \;\Longrightarrow\; x = 0 \;\Longrightarrow\; \text{``inflection at } (0,0)\text{''} \]
Solve for the candidate, then test the sign on both sides before you claim anything.
\[ f''(x) = 12x^2 \]
Test a point on each side of the candidate
Why: Twelve times negative one squared is 12; twelve times one squared is also 12. Both positive.
\[ f''(-1) = 12 > 0, \qquad f''(1) = 12 > 0 \]
Conclude: no sign change, so no inflection point
Why: The curve is concave up on both sides. It flattens at the origin without ever switching its bend - the classic bowl shape of a fourth-power graph.
\[ x^4 \text{ is concave up on } (-\infty,0) \text{ and on } (0,\infty) \]
Picture it
Figure (svg): Number line split at x equals 0 and x equals 2, showing plus, minus, plus from left to right
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:
Find every inflection point of this function.
Worked example
Find every inflection point of this function.
\[ f(x) = x^4 - 4x^3 \]
Differentiate twice
Why: Power rule twice, term by term.
\[ f'(x) = 4x^3 - 12x^2, \qquad f''(x) = 12x^2 - 24x \]
Factor the second derivative
Why: Factored form hands you the candidates and the signs at the same time.
\[ f''(x) = 12x(x-2) \]
Read off the candidates
Why: A product is zero exactly when one factor is zero; the polynomial is defined everywhere, so there are no undefined points to add.
\[ x = 0 \quad \text{and} \quad x = 2 \]
Test one value in each of the three intervals
Why: Use negative one, one, and three. Negative one gives 12 plus 24, one gives 12 minus 24, three gives 108 minus 72.
\[ f''(-1) = 36 > 0, \quad f''(1) = -12 < 0, \quad f''(3) = 36 > 0 \]
Figure (svg): Number line split at x equals 0 and x equals 2, showing plus, minus, plus from left to right
Both candidates survive, so substitute back for the y-coordinates
Why: The sign flips at each candidate. Now evaluate the original function to name the actual points.
\[ f(0) = 0, \qquad f(2) = 16 - 32 = -16 \]
\[ \text{inflection points: } (0,\,0) \text{ and } (2,\,-16) \]
Verify with a second pair of test points
Why: At x equals negative one half the second derivative is 3 plus 12, which is 15, still positive. At x equals five halves it is 75 minus 60, which is 15, still positive. And at x equals one and a half it is 27 minus 36, which is negative 9. The chart holds up.
\[ f''\left(-\tfrac{1}{2}\right) = 15, \quad f''\left(\tfrac{3}{2}\right) = -9, \quad f''\left(\tfrac{5}{2}\right) = 15 \]
Blank canvas
Draw it
Draw what Worked example: inflection points of a quartic just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.
Concept
Figure (svg): An S-shaped cube-root curve passing through the origin with a vertical tangent, concave up on the left and concave down on the right
Candidates are not only the zeros. The cube-root function is the standard example.
\[ f(x) = x^{1/3} \]
\[ f'(x) = \tfrac{1}{3}x^{-2/3}, \qquad f''(x) = -\tfrac{2}{9}x^{-5/3} \]
The second derivative does not exist at the origin. But the function itself is perfectly well defined there, so the origin is still a point on the graph.
\[ f''(-1) = \tfrac{2}{9} > 0, \qquad f''(1) = -\tfrac{2}{9} < 0 \]
Concave up on the left, concave down on the right: the concavity changes, so the origin is an inflection point - one with a vertical tangent line.
Concept
A vertical asymptote also splits the number line, and the concavity very often differs on its two sides.
That is not an inflection point. There is no point there at all - the function has no value to plot.
Always intersect your candidate list with the domain before reporting anything.
Picture it
Animation
Shows: No point, no inflection point — a rendered Manim animation.
Rendered with Manim.
Takeaway: x to the fourth has a zero second derivative at the origin and no inflection.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The reasoning: the concavity is different on the two sides of the asymptote, so the bend switched, so that must be an inflection point.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The numerator is always positive, so the sign is the sign of the cube of x squared minus one - negative just inside, positive just outside.
Check the domain first. The denominator vanishes at both of those x-values.
Why: The numerator is always positive, so the sign is the sign of the cube of x squared minus one - negative just inside, positive just outside. The flip is real; the conclusion is not.
Trap
The reasoning: the concavity is different on the two sides of the asymptote, so the bend switched, so that must be an inflection point.
\[ f(x) = \frac{x^2}{x^2-1}, \qquad f''(x) = \frac{6x^2+2}{(x^2-1)^3} \]
Note the sign flip at x equals one and call it an inflection point
Why: The numerator is always positive, so the sign is the sign of the cube of x squared minus one - negative just inside, positive just outside. The flip is real; the conclusion is not.
\[ \text{``inflection points at } x = -1 \text{ and } x = 1\text{''} \]
Check the domain first. The denominator vanishes at both of those x-values.
\[ x^2 - 1 = 0 \;\Longrightarrow\; x = \pm 1 \;\text{ excluded from the domain} \]
Discard both candidates: they are not on the graph
Why: An inflection point is a point of the curve. At these x-values there is no curve - there is a vertical asymptote and a gap.
Look for zeros of the second derivative inside the domain
Why: Six x squared plus two is at least 2 for every real x, so the numerator is never zero. There is nowhere left for an inflection point to live.
\[ 6x^2 + 2 \ge 2 > 0 \;\Longrightarrow\; \text{this function has no inflection points} \]
Translation
\( f(x) = \frac{x^2}{x^2-1}, \qquad f''(x) = \frac{6x^2+2}{(x^2-1)^3} \)
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.
Prediction
Predict first
At which x-values does the graph of f have an inflection point?
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: At x = 3 only
Why: The factor x squared is positive on both sides of zero, so the sign of the second derivative is the sign of x minus 3 everywhere except at the single point x equals 0. That makes it negative on the whole interval to the left of 3 and positive to the right, so the only sign change is at x equals 3.
Check
A function is defined and twice differentiable everywhere, and its second derivative is given below. Build the sign chart first.
\[ f''(x) = x^2(x-3) \]
Check your understanding
At which x-values does the graph of f have an inflection point?
Answer: A
Why: The factor x squared is positive on both sides of zero, so the sign of the second derivative is the sign of x minus 3 everywhere except at the single point x equals 0. That makes it negative on the whole interval to the left of 3 and positive to the right, so the only sign change is at x equals 3.
Section
Part 3
Concept
Suppose you already found a critical point where the first derivative is zero. The tangent line there is flat. Is it a peak or a valley?
Look at the bend. A flat tangent at the bottom of a bowl is a minimum. A flat tangent at the top of a dome is a maximum.
That single observation is the Second Derivative Test.
Intuition
Figure (svg): A bowl-shaped curve with a flat tangent at its lowest point beside a dome-shaped curve with a flat tangent at its highest point
Both pictures have exactly the same first-derivative information: the tangent line is horizontal.
Only the concavity distinguishes them. Holds water at a flat spot means you are at the bottom; spills water at a flat spot means you are at the top.
So you do not need to test values on either side. One number - the second derivative at the point - settles it.
Concept
Suppose the second derivative is continuous near a critical number where the first derivative is zero.
\[ f'(c) = 0 \;\text{ and }\; f''(c) > 0 \;\Longrightarrow\; f \text{ has a local minimum at } c \]
\[ f'(c) = 0 \;\text{ and }\; f''(c) < 0 \;\Longrightarrow\; f \text{ has a local maximum at } c \]
\[ f'(c) = 0 \;\text{ and }\; f''(c) = 0 \;\Longrightarrow\; \text{no conclusion} \]
Sorting
Sort into buckets
These are the pieces of Concavity, Inflection Points, and Curve Sketching, out of order. Put each one back under the part of the lesson it belongs to.
Picture it
Animation
Shows: The second derivative test with its inconclusive case highlighted.
A zero second derivative decides nothing.
Takeaway: A negative second derivative means a maximum, positive means a minimum, and zero means the test tells you nothing — fall back to the first derivative's sign.
Concept
First: the test only applies at a point where the first derivative is zero. If the derivative is merely undefined there - a corner or a cusp - the test says nothing and you use the First Derivative Test.
Second: a positive second derivative means minimum, not maximum. Positive sounds like up, and students pair up with maximum. Go back to the bowl picture instead of trusting the word.
Ranking
Put in order
Put the moves of Worked example: classifying with the Second Derivative Test 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 numbers of a polynomial come only from the derivative being zero, since a polynomial derivative is defined everywhere.
Worked example
Find and classify the critical points.
\[ f(x) = x^3 - 12x \]
Differentiate and set equal to zero
Why: Critical numbers of a polynomial come only from the derivative being zero, since a polynomial derivative is defined everywhere.
\[ f'(x) = 3x^2 - 12 = 3(x-2)(x+2) = 0 \]
Read off the critical numbers
Why: A product is zero when a factor is zero.
\[ x = -2 \quad \text{and} \quad x = 2 \]
Differentiate again
Why: One more power-rule pass; this is the tool that will classify both points.
\[ f''(x) = 6x \]
Evaluate the second derivative at each critical number
Why: Six times negative two is negative twelve; six times two is twelve.
\[ f''(-2) = -12 < 0, \qquad f''(2) = 12 > 0 \]
Classify and compute the values
Why: Negative means concave down means a peak; positive means concave up means a valley. Substitute into the original function for the heights.
\[ \text{local max } (-2,\,16), \qquad \text{local min } (2,\,-16) \]
Verify by comparing nearby function values
Why: Near the claimed maximum: f of negative three is 9 and f of negative one is 11, both below 16. Near the claimed minimum: f of one is negative 11 and f of three is negative 9, both above negative 16. The classification holds.
| x | -3 | -2 | -1 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| f(x) | 9 | 16 | 11 | -11 | -16 | -9 |
Picture it
Animation
Shows: Each line of the worked example "classifying with the Second Derivative Test", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Near the claimed maximum: f of negative three is 9 and f of negative one is 11, both below 16. Near the claimed minimum: f of one is negative 11 and f of three is negative 9, both above negative 16. The classification holds.
Concept
When the second derivative is zero at the critical point, the test gives you nothing. It does not say there is no extremum.
Compare it to a metal detector that beeps for iron and stays quiet for aluminium. Silence is not proof the ground is empty.
The fix is always the same: go back to the First Derivative Test and check the sign of the first derivative on each side.
Concept
All three of these have a zero first derivative and a zero second derivative at the origin, and all three behave differently there.
| function | first derivative | second derivative | what actually happens at the origin |
|---|---|---|---|
| x to the fourth | 4x cubed | 12x squared | local minimum |
| negative x to the fourth | negative 4x cubed | negative 12x squared | local maximum |
| x cubed | 3x squared | 6x | neither - an inflection point with a flat tangent |
One reading of zero, three different truths. That is precisely why the test has to refuse to answer.
Trade off
Comparison matrix
From Three functions, same inconclusive reading: every row here is a choice with a cost. Fill the second derivative column, then say which row you would actually pick and what you give up for it.
| function | first derivative | second derivative | what actually happens at the origin |
|---|---|---|---|
| x to the fourth | 4x cubed | 12x squared | local minimum |
| negative x to the fourth | negative 4x cubed | negative 12x squared | local maximum |
| x cubed | 3x squared | 6x | neither - an inflection point with a flat tangent |
Anomaly
Predict first
A student writes this, and it looks reasonable:
The reasoning: the test came back zero, so there is no maximum and no minimum here.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This treats a refusal to answer as an answer of no.
Inconclusive means switch tools. Use the First Derivative Test.
Why: This treats a refusal to answer as an answer of no. The Second Derivative Test never has the authority to rule an extremum out.
Trap
The reasoning: the test came back zero, so there is no maximum and no minimum here.
\[ f(x) = x^4, \qquad f'(0) = 0, \qquad f''(0) = 0 \]
Report no local extremum at the origin
Why: This treats a refusal to answer as an answer of no. The Second Derivative Test never has the authority to rule an extremum out.
\[ \text{``no local extremum at } x=0\text{''} \quad \text{(false)} \]
Inconclusive means switch tools. Use the First Derivative Test.
\[ f'(x) = 4x^3 \]
Check the sign of the first derivative on each side of zero
Why: Four times negative one cubed is negative four; four times one cubed is positive four.
\[ f'(-1) = -4 < 0, \qquad f'(1) = 4 > 0 \]
Decreasing then increasing means a local minimum
Why: The function falls into the point and climbs back out of it, which is the definition of a valley.
| x | -1 | -0.5 | 0 | 0.5 | 1 |
|---|---|---|---|---|---|
| f(x) | 1 | 0.0625 | 0 | 0.0625 | 1 |
\[ \text{local (in fact absolute) minimum at } (0,\,0) \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Four times negative one cubed is negative four; four times one cubed is positive four.
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:
This treats a refusal to answer as an answer of no. The Second Derivative Test never has the authority to rule an extremum out.
Pattern
The First Derivative Test is the reliable one; the Second Derivative Test is the fast one. Use speed where it is safe.
Commit first
Predict first
What does the Second Derivative Test say about the critical point at x = 4?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: Local minimum
Why: The first derivative is 3x squared minus 12x, which factors as 3x times x minus 4, so x equals 4 is critical. The second derivative is 6x minus 12, and at x equals 4 that is 24 minus 12, which equals 12. A positive second derivative means concave up, so the flat tangent sits at the bottom of a bowl: a local minimum.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Find the derivative, confirm the point is critical, then classify it.
\[ f(x) = x^3 - 6x^2 + 5 \]
Check your understanding
What does the Second Derivative Test say about the critical point at x = 4?
Answer: A
Why: The first derivative is 3x squared minus 12x, which factors as 3x times x minus 4, so x equals 4 is critical. The second derivative is 6x minus 12, and at x equals 4 that is 24 minus 12, which equals 12. A positive second derivative means concave up, so the flat tangent sits at the bottom of a bowl: a local minimum.
Check
Both derivatives vanish at the origin for this function.
\[ f(x) = x^4, \qquad f'(0) = 0, \qquad f''(0) = 0 \]
Check your understanding
What is the correct conclusion at x = 0?
Answer: A
Why: A zero second derivative at a critical point tells you nothing, so you switch tests. The first derivative 4x cubed is negative to the left of zero and positive to the right, so the function decreases into the origin and increases out of it: a local minimum, with value 0.
Section
Part 4
Concept
A graphing calculator draws a picture. Your job is to draw a picture you can defend - every feature justified by a computation.
Hit all five and the drawing has almost no freedom left. That is what makes the method reliable.
Picture it
Animation
Shows: A sketch is an argument — a rendered Manim animation.
Rendered with Manim.
Takeaway: If you cannot name the reason for a bump, it does not go on the page.
Picture it
Animation
Shows: The curve-sketching checklist in order.
Work the list and the shape has nowhere to hide.
Takeaway: Domain and asymptotes, then the first derivative, then the second. Worked in order, the checklist leaves the curve no freedom.
Concept
Stack the first-derivative sign chart above the second-derivative sign chart, using the same tick marks for both.
| sign of the first derivative | sign of the second derivative | the shape of that piece |
|---|---|---|
| plus | plus | rising, holding water - the bottom-left of a bowl |
| plus | minus | rising, spilling water - the top-left of a dome |
| minus | minus | falling, spilling water - the top-right of a dome |
| minus | plus | falling, holding water - the bottom-right of a bowl |
Each interval gets one of four little arcs. Draw the arcs in order, join them, and you have the curve.
Constraint
Discussion prompt
Run Pattern: the curve-sketching checklist with this step confiscated:
Intercepts. Set the function to zero for the x-intercepts; evaluate at zero for the y-intercept.
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
Edge cases
Discussion prompt
Pattern: the curve-sketching checklist 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:
Step zero
Discussion prompt
Full sketch, part 1: gathering the data — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Domain and intercepts
Answer:
Worked example
Sketch this curve completely.
\[ f(x) = x^4 - 4x^3 \]
Domain and intercepts
Why: A polynomial is defined for every real number. Factor out the largest common power to find where it crosses the horizontal axis.
\[ x^4 - 4x^3 = x^3(x-4) = 0 \;\Longrightarrow\; x = 0,\; x = 4 \]
Symmetry and end behaviour
Why: Replacing x by its negative gives x to the fourth plus four x cubed, which is neither the original nor its opposite, so there is no symmetry. Even degree with a positive leading coefficient sends both ends upward.
\[ \lim_{x \to \infty} f(x) = \infty, \qquad \lim_{x \to -\infty} f(x) = \infty \]
First derivative and critical numbers
Why: Factor out the common four x squared; the derivative of a polynomial is never undefined, so zeros are the only critical numbers.
\[ f'(x) = 4x^3 - 12x^2 = 4x^2(x-3) \;\Longrightarrow\; x = 0,\; x = 3 \]
Build the first-derivative sign chart
Why: The four x squared factor is positive on both sides of zero, so only the factor x minus 3 controls the sign. Negative left of 3, positive right of 3.
\[ f'(-1) = -16, \quad f'(1) = -8, \quad f'(4) = 64 \]
Read the extrema off the chart
Why: The sign switches from negative to positive only at x equals 3, so that is a local minimum. At x equals 0 the derivative touches zero without switching sign, so there is no extremum there - just a flat spot on the way down.
\[ \text{local min at } (3,\,-27); \quad x = 0 \text{ is a critical point but not an extremum} \]
Verify the two derivative values that matter most
Why: At x equals 3 the derivative is four times nine times zero, which is 0, confirming the critical point. At x equals 3 the function value is 81 minus 108, which is negative 27, confirming the height.
\[ f'(3) = 4(9)(0) = 0, \qquad f(3) = 81 - 108 = -27 \]
Worked example
Same function. Now add the concavity information and draw.
\[ f(x) = x^4 - 4x^3, \qquad f''(x) = 12x(x-2) \]
Reuse the second-derivative sign chart
Why: We already found it: positive, negative, positive across the cut points 0 and 2, with inflection points at both.
\[ \text{CU on } (-\infty,0),\; \text{CD on } (0,2),\; \text{CU on } (2,\infty) \]
Cross-check the local minimum with the Second Derivative Test
Why: Twelve times three times one is 36, positive, so the curve is concave up at the critical point - a valley, exactly as the first-derivative sign chart said.
\[ f''(3) = 36 > 0 \]
| interval | first derivative | second derivative | shape |
|---|---|---|---|
| left of 0 | minus | plus | falling, holding water |
| 0 to 2 | minus | minus | falling, spilling water |
| 2 to 3 | minus | plus | falling, holding water |
| right of 3 | plus | plus | rising, holding water |
Plot the anchors and connect them
Why: The anchors are the intercepts at 0 and 4, the inflection points at the origin and at the point two, negative sixteen, and the minimum at three, negative twenty-seven.
Figure (svg): Sketch of the quartic x to the fourth minus four x cubed: falling from upper left, a flat spot at the origin, a minimum near x equals 3, then rising steeply through x equals 4
Verify with a table of plotted values
Why: The values fall steadily from 5 down to negative 27 and then climb back through zero at x equals 4, exactly matching a curve that decreases until 3 and increases after. The flat spot at the origin shows up as the tiny drop from 0 to negative 3 across a whole unit.
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|
| f(x) | 5 | 0 | -3 | -16 | -27 | 0 | 125 |
Discrimination
Sort into buckets
Sort these by first derivative, from memory, without looking back at Full sketch, part 2: bending and drawing. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Intuition
Nothing in that sketch was artistic judgement. Each interval had a forced shape, and the anchor points pinned the heights.
If someone hands you a graph and asks where the mistake is, this is how you find it: pick an interval, read the two signs, and check the drawn arc matches.
Concept
Draw the vertical and horizontal asymptotes as dashed lines first. They box the plane into regions, and each branch of the curve lives inside one region.
Then figure out which way each branch runs off toward its vertical asymptote, using the sign of the function just to the left and just to the right.
Ranking
Put in order
Put the moves of Rational sketch, part 1: domain, symmetry, asymptotes 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 fraction is undefined exactly where its denominator is zero.
Worked example
Sketch this rational function completely.
\[ f(x) = \frac{x^2}{x^2-1} \]
Find the domain
Why: A fraction is undefined exactly where its denominator is zero.
\[ x^2 - 1 = 0 \;\Longrightarrow\; x = \pm 1 \quad \text{(excluded)} \]
Intercepts
Why: A fraction is zero only when its numerator is zero, and x squared is zero only at the origin. Substituting zero gives zero over negative one.
\[ \text{only intercept: } (0,\,0) \]
Test for symmetry
Why: Replacing x by its negative squares away both minus signs, so the function is unchanged. That means the graph is a mirror image across the vertical axis - draw half of it and reflect.
\[ f(-x) = \frac{(-x)^2}{(-x)^2-1} = \frac{x^2}{x^2-1} = f(x) \]
Vertical asymptotes and the sign on each side
Why: Just inside the interval between the asymptotes the denominator is a small negative number, so the fraction dives downward; just outside it is a small positive number, so the fraction shoots upward.
\[ \lim_{x \to 1^{-}} f(x) = -\infty, \qquad \lim_{x \to 1^{+}} f(x) = +\infty \]
Horizontal asymptote
Why: Numerator and denominator have the same degree, so divide both by x squared: the limit is one over one minus a vanishing term.
\[ \lim_{x \to \pm\infty} \frac{x^2}{x^2-1} = \lim_{x \to \pm\infty} \frac{1}{1 - \frac{1}{x^2}} = 1 \]
Verify the asymptote claims with actual numbers
Why: At x equals one and one tenth the value is 1.21 over 0.21, about 5.76 - large and positive, as predicted. At x equals nine tenths it is 0.81 over negative 0.19, about negative 4.26. And at x equals 10 the value is 100 over 99, about 1.01, hugging the horizontal asymptote from above.
| x | 0.9 | 1.1 | 3 | 10 |
|---|---|---|---|---|
| f(x) | -4.26 | 5.76 | 1.125 | 1.0101 |
Worked example
Same function. Now the two derivatives.
\[ f(x) = \frac{x^2}{x^2-1} \]
Quotient rule for the first derivative
Why: Low times the derivative of high minus high times the derivative of low, all over low squared. The x cubed terms cancel, which is why the result is so clean.
\[ f'(x) = \frac{2x(x^2-1) - x^2(2x)}{(x^2-1)^2} = \frac{-2x}{(x^2-1)^2} \]
Read the first-derivative sign chart
Why: The denominator is a square, so it is positive everywhere it is defined. The sign is therefore just the sign of negative two x: positive for negative x, negative for positive x.
\[ \text{increasing on } (-\infty,-1) \text{ and } (-1,0); \quad \text{decreasing on } (0,1) \text{ and } (1,\infty) \]
Differentiate again
Why: Write the first derivative as negative two x times the denominator to the negative two power, use the product and chain rules, then factor out the denominator to the negative three power.
\[ f''(x) = \frac{-2(x^2-1) + 8x^2}{(x^2-1)^3} = \frac{6x^2+2}{(x^2-1)^3} \]
Read the concavity
Why: The numerator is at least 2 for every real x, so the sign is the sign of the cube of x squared minus one: negative between the asymptotes, positive outside them. Since the sign only flips at excluded points, there are no inflection points.
\[ \text{CD on } (-1,1); \quad \text{CU on } (-\infty,-1) \text{ and } (1,\infty) \]
Classify the one critical point and draw
Why: The second derivative at zero is two over negative one, which is negative two - concave down, so the origin is a local maximum. Three branches: two outside the asymptotes falling toward the line at height one from above, and one arch between them peaking at the origin.
Figure (svg): Sketch of the rational function x squared over x squared minus one: two outer branches above the horizontal asymptote at height one, and a middle arch peaking at the origin and diving down at both vertical asymptotes
Verify the local maximum against nearby values
Why: At x equals one half the value is 0.25 over negative 0.75, which is about negative 0.333, and at x equals negative one half it is the same. Both sit below the value 0 at the origin, so the origin really is the top of that middle arch.
\[ f\left(-\tfrac{1}{2}\right) = f\left(\tfrac{1}{2}\right) = -\tfrac{1}{3} < 0 = f(0) \]
Blank canvas
Draw it
Draw what Rational sketch, part 2: derivatives and the drawing just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.
Prediction
Predict first
How many inflection points does this graph have?
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: None
Why: The numerator six x squared plus two is at least 2 for every real x, so the second derivative is never zero. It is undefined only at x equals negative one and x equals one, and those are not in the domain, so no point of the graph is a candidate at all.
Check
Use the second derivative you just computed, and remember to intersect the candidates with the domain.
\[ f(x) = \frac{x^2}{x^2-1}, \qquad f''(x) = \frac{6x^2+2}{(x^2-1)^3} \]
Check your understanding
How many inflection points does this graph have?
Answer: A
Why: The numerator six x squared plus two is at least 2 for every real x, so the second derivative is never zero. It is undefined only at x equals negative one and x equals one, and those are not in the domain, so no point of the graph is a candidate at all.
Picture it
Figure (svg): Sketch of x times e to the negative x: rising from below the axis through the origin to a peak near x equals one, then decaying toward the horizontal axis with an inflection near x equals two
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:
Analyze and sketch this function, which shows up in probability and in circuit discharge.
Worked example
Analyze and sketch this function, which shows up in probability and in circuit discharge.
\[ f(x) = x e^{-x} \]
Differentiate with the product rule
Why: The derivative of x is 1 and the derivative of the exponential is negative itself; factor the exponential out of both terms.
\[ f'(x) = e^{-x} - x e^{-x} = e^{-x}(1-x) \]
Find the critical number
Why: An exponential is never zero, so the only way the product is zero is one minus x equals zero.
\[ e^{-x}(1-x) = 0 \;\Longrightarrow\; x = 1 \]
Differentiate again
Why: Product rule on the exponential times one minus x: the exponential brings down a minus sign, and the second factor differentiates to negative one. Collect terms.
\[ f''(x) = -e^{-x}(1-x) - e^{-x} = e^{-x}(x-2) \]
Classify the critical point with the Second Derivative Test
Why: At x equals 1 the second factor is negative one, and the exponential is positive, so the whole thing is negative - concave down, hence a peak.
\[ f''(1) = -e^{-1} < 0 \;\Longrightarrow\; \text{local max at } \left(1,\; \tfrac{1}{e}\right) \]
Find the inflection point
Why: The exponential is always positive, so the sign of the second derivative is the sign of x minus 2: negative to the left, positive to the right. A genuine sign change.
\[ \text{inflection at } \left(2,\; \tfrac{2}{e^2}\right) \approx (2,\, 0.271) \]
Add the end behaviour and draw
Why: As the input grows without bound the exponential decay beats the linear growth, so the curve settles onto the horizontal axis from above; in the negative direction both factors push it downward without bound.
Figure (svg): Sketch of x times e to the negative x: rising from below the axis through the origin to a peak near x equals one, then decaying toward the horizontal axis with an inflection near x equals two
Verify the maximum numerically
Why: The claimed peak value is about 0.3679. Just left, at nine tenths, the value is about 0.3659; just right, at one and one tenth, it is about 0.3662. Both are smaller, so the point really is a local maximum.
| x | 0.9 | 1 | 1.1 | 2 |
|---|---|---|---|---|
| f(x) | 0.3659 | 0.3679 | 0.3662 | 0.2707 |
Notation
Annotate
From Worked example: an exponential curve — read this one piece at a time. What is each part doing?
On: \( \text{inflection at } \left(2,\; \tfrac{2}{e^2}\right) \approx (2,\, 0.271) \)
Section
Part 5
Concept
When the picture in front of you is the derivative, every question about the original function becomes a question about the picture's height and slope.
| feature of the picture (the derivative) | what it says about the original function |
|---|---|
| above the horizontal axis | the function is increasing |
| below the horizontal axis | the function is decreasing |
| crosses the axis going up | local minimum of the function |
| crosses the axis going down | local maximum of the function |
| touches the axis without crossing | flat spot, but no extremum |
Height of the derivative equals slope of the function. That one sentence powers the whole table.
Comparison
Comparison matrix
From What a graph of the first derivative tells you: refill the what it says about the original function column from what you know. The rest of the table is as it appeared.
| feature of the picture (the derivative) | what it says about the original function |
|---|---|
| above the horizontal axis | the function is increasing |
| below the horizontal axis | the function is decreasing |
| crosses the axis going up | local minimum of the function |
| crosses the axis going down | local maximum of the function |
| touches the axis without crossing | flat spot, but no extremum |
Picture it
Animation
Shows: What a graph of the first derivative tells you — a rendered Manim animation.
Rendered with Manim.
Takeaway: Where this crosses zero, the original has a flat tangent.
Concept
Now look at the slope of the derivative's graph rather than its height.
| feature of the picture (the derivative) | what it says about the original function |
|---|---|
| rising | the original function is concave up |
| falling | the original function is concave down |
| a peak or a valley of the picture | an inflection point of the original function |
So a turning point of the derivative is never an extremum of the function. It is where the bend switches.
Trade off
Comparison matrix
From The slope of the derivative graph is the concavity: every row here is a choice with a cost. Fill the what it says about the original function column, then say which row you would actually pick and what you give up for it.
| feature of the picture (the derivative) | what it says about the original function |
|---|---|
| rising | the original function is concave up |
| falling | the original function is concave down |
| a peak or a valley of the picture | an inflection point of the original function |
Picture it
Animation
Shows: The slope of the derivative graph is the concavity — a rendered Manim animation.
Rendered with Manim.
Takeaway: One chain of equivalences, read in either direction.
Intuition
Picture the three graphs stacked one above another, sharing the same horizontal axis, and slide a single vertical line across all three.
Where the middle graph crosses zero, the top graph levels out. Where the bottom graph crosses zero, the middle graph turns around and the top graph changes its bend.
Every question of this type is answered by that sliding line. Read down one column, never sideways between graphs.
Explain it
Discussion prompt
Explain Three graphs, one vertical line 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:
Where the middle graph crosses zero, the top graph levels out. Where the bottom graph crosses zero, the middle graph turns around and the top graph changes its bend.
Concept
Keep this in your head for exams. It is the same information three ways.
| at a point where | the first derivative | the second derivative | the function |
|---|---|---|---|
| there is a local max | is zero or undefined | is negative or zero | peaks |
| there is a local min | is zero or undefined | is positive or zero | bottoms out |
| there is an inflection | has a peak or valley | changes sign | switches bend |
| the function is steepest downhill | has a minimum | is zero | is inflecting |
The last row is the one that catches people, and it is exactly the trap coming up next.
Discrimination
Sort into buckets
Sort these by the first derivative, from memory, without looking back at The translation table, both directions. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Ranking
Put in order
Put the moves of Worked example: describing a function from its derivative 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. Those are the critical numbers of the original function - the places where its tangent line goes flat.
Worked example
Figure (svg): An upward-opening parabola crossing the horizontal axis at x equals negative two and x equals two, with its vertex at the point zero, negative four
The curve shown is the derivative, not the function. Describe the function completely.
\[ f'(x) = x^2 - 4 \]
Find where the derivative is zero
Why: Those are the critical numbers of the original function - the places where its tangent line goes flat.
\[ x^2 - 4 = 0 \;\Longrightarrow\; x = -2, \; x = 2 \]
Read the sign of the derivative from the picture
Why: The parabola is above the axis outside the two roots and below it in between.
\[ f' > 0 \text{ on } (-\infty,-2) \cup (2,\infty); \quad f' < 0 \text{ on } (-2,2) \]
Classify the two critical points
Why: The derivative goes from positive to negative at negative two, so the function peaks there; it goes from negative to positive at two, so the function bottoms out there.
\[ \text{local max at } x=-2, \qquad \text{local min at } x=2 \]
Get the concavity from the slope of the picture
Why: The parabola falls until its vertex and rises after, so the second derivative is negative then positive. Algebraically the derivative of x squared minus four is two x.
\[ f''(x) = 2x \;\Longrightarrow\; \text{CD on } (-\infty,0), \; \text{CU on } (0,\infty) \]
Name the inflection point of the function
Why: The vertex of the picture is at x equals zero, and the second derivative changes sign there, so the original function inflects at that input.
\[ \text{inflection of } f \text{ at } x = 0 \]
Verify with an actual antiderivative
Why: One function with this derivative is x cubed over three minus four x. Its values are 3, 5.33, 3.67 at x equal to negative three, negative two, negative one, so negative two really is a local max; and negative 3.67, negative 5.33, negative 3 at one, two, three, so two really is a local min.
\[ f(x) = \tfrac{x^3}{3} - 4x \;\Rightarrow\; f(-2) = \tfrac{16}{3}, \quad f(2) = -\tfrac{16}{3} \]
Picture it
Animation
Shows: Each line of the worked example "describing a function from its derivative", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: One function with this derivative is x cubed over three minus four x. Its values are 3, 5.33, 3.67 at x equal to negative three, negative two, negative one, so negative two really is a local max; and negative 3.67, negative 5.33, negative 3 at one, two, three, so two really is a local min.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The reasoning: the picture has its lowest point at the origin, so the function has a minimum at the origin.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This treats the plotted curve as the function itself.
Ask what the height of the picture means before you look at its shape.
Why: This treats the plotted curve as the function itself. It is a very natural slip, because a valley in a picture screams minimum.
Trap
The reasoning: the picture has its lowest point at the origin, so the function has a minimum at the origin.
\[ \text{picture shown: } y = f'(x) = x^2 - 4 \]
Report a local minimum of f at x equals zero
Why: This treats the plotted curve as the function itself. It is a very natural slip, because a valley in a picture screams minimum.
\[ f'(0) = -4 \neq 0 \;\Longrightarrow\; x=0 \text{ is not even a critical number} \]
Ask what the height of the picture means before you look at its shape.
\[ \text{height of the picture} = \text{slope of } f \]
At x equals zero the height is negative four
Why: So the function has slope negative four there. It is falling steeply, not sitting at a minimum.
A valley in the derivative means the slope stops decreasing and starts increasing
Why: That is the second derivative changing sign from negative to positive, so the function has an inflection point at that input - its steepest downhill moment.
\[ \text{valley of } f' \;\Longrightarrow\; \text{inflection point of } f, \text{ not an extremum} \]
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.
Pattern
Then answer the question by reading your annotations, never by looking at the shape of the curve as a whole.
Real world
Discussion prompt
Outside this lesson: where does Concavity, Inflection Points, and Curve Sketching actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: reading a derivative graph is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck explains what the second derivative measures and distinguishes concave up from concave down. It covers inflection points and the sign change they require, the Second Derivative Test and its inconclusive case, and first- and second-derivative sign charts, then works the full curve-sketching checklist end to end on a polynomial, a rational function, and an exponential. It targets the classic errors: calling every zero of the second derivative an inflection point, confusing decreasing with concave down, treating an inconclusive test as proof that there is no extremum, and reading features of a function off the graph of its derivative.
Picture it
Figure (svg): An upward-opening parabola crossing the horizontal axis at x equals one and x equals three, with its lowest point at x equals two
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:
The curve drawn below is the derivative of a function that is twice differentiable everywhere.
Check
The curve drawn below is the derivative of a function that is twice differentiable everywhere.
\[ f'(x) = (x-1)(x-3) \]
Figure (svg): An upward-opening parabola crossing the horizontal axis at x equals one and x equals three, with its lowest point at x equals two
Check your understanding
What is true about the original function f at x = 2?
Answer: A
Why: The picture is the derivative, and it has its lowest point at x equals 2. A valley of the derivative is where the second derivative changes sign from negative to positive, which is an inflection point of f. Directly: the second derivative is 2x minus 4, which is negative before 2 and positive after.
Concept
When a quantity is described as growing but slowing down, that is a positive first derivative and a negative second derivative. Concave down.
Economists call this diminishing returns: the tenth employee adds less than the ninth did. The total still climbs; the climb flattens.
The reverse pairing - growing and speeding up - is what people mean by explosive or exponential growth. Same direction, opposite bend.
Analogy
Discussion prompt
Explain Concavity in the real world: the rate of the rate 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:
When a quantity is described as growing but slowing down, that is a positive first derivative and a negative second derivative. Concave down.
Picture it
Animation
Shows: Concavity in the real world — a rendered Manim animation.
Rendered with Manim.
Takeaway: Journalists call this a slowdown. It is still growth.
Intuition
During an epidemic, the headline number is the cumulative case count. It rises the whole way through, so its first derivative is positive throughout - that never tells you anything encouraging.
The number people actually watch for is the day the daily count peaks. That is the inflection point of the cumulative curve: still rising, but from that day on, rising more slowly.
This is why inflection points get their own name in news stories. They are the earliest honest sign that something has turned.
Counterexample
Discussion prompt
During an epidemic, the headline number is the cumulative case count. It rises the whole way through, so its first derivative is positive throughout - that never tells you anything encouraging.
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:
The number people actually watch for is the day the daily count peaks. That is the inflection point of the cumulative curve: still rising, but from that day on, rising more slowly.
Elimination
Eliminate the wrong options
What do the first and second derivatives of the revenue function do over this stretch?
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: Revenue rising every month means the rate of change is positive, so the first derivative is positive. The monthly increases shrinking means that rate of change is itself decreasing, so its derivative - the second derivative - is negative. Rising and concave down.
Check
A shop's monthly revenue is higher every month than the month before, but each month's increase is smaller than the previous month's increase.
Check your understanding
What do the first and second derivatives of the revenue function do over this stretch?
Answer: A
Why: Revenue rising every month means the rate of change is positive, so the first derivative is positive. The monthly increases shrinking means that rate of change is itself decreasing, so its derivative - the second derivative - is negative. Rising and concave down.
Concept
| question | the tool | the deciding sign |
|---|---|---|
| Is the curve rising or falling here? | first derivative | positive means rising |
| Is it a peak or a valley? | Second Derivative Test at a critical number | negative means peak |
| Does it hold water or spill it? | second derivative | positive means holds water |
| Is this an inflection point? | sign chart of the second derivative | the sign must actually flip |
| Is that candidate even eligible? | the domain | the point must be on the graph |
Five questions, five tools. Everything else in this deck is practice at running them in order.
Comparison
Comparison matrix
From One page of everything: refill the the deciding sign column from what you know. The rest of the table is as it appeared.
| question | the tool | the deciding sign |
|---|---|---|
| Is the curve rising or falling here? | first derivative | positive means rising |
| Is it a peak or a valley? | Second Derivative Test at a critical number | negative means peak |
| Does it hold water or spill it? | second derivative | positive means holds water |
| Is this an inflection point? | sign chart of the second derivative | the sign must actually flip |
| Is that candidate even eligible? | the domain | the point must be on the graph |
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — What the Second Derivative Measures · Inflection Points · The Second Derivative Test · The Full Curve-Sketching Checklist · Reading f, f prime, and f double prime Off One Another. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
| If you see | Conclude |
|---|---|
| second derivative positive | concave up on that interval |
| second derivative zero with no sign change | not an inflection point |
| derivative zero and second derivative negative | local maximum |
| derivative zero and second derivative zero | inconclusive - use the First Derivative Test |
| a valley in the graph of the derivative | an inflection point of the function |
Next up: optimization, where this whole toolkit gets pointed at a single real-world quantity you want as large or as small as possible.
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