Absolute and local extrema and the difference between them, the Extreme Value Theorem and its two hypotheses, Fermat's theorem that an interior extremum forces a zero derivative, critical points including those where the derivative fails to exist, and the closed-interval method for finding absolute extrema.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
Maxima and Minima
Objectives
Five outcomes. The last is a complete method, and it works because the first four have narrowed the search to three kinds of place.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 317-328 — the section these objectives are drawn from
Warm-up
Section 3.2 sketched derivatives from graphs and noticed that a turning point sits above a zero of the derivative — but also that a zero derivative need not be a turning point.
Discussion prompt
The cubing function has a zero derivative at the origin and no turning point there. What extra condition does a genuine maximum or minimum need?
Hint: Look at what the derivative's sign does on either side.
Answer:
\[ f(x) = x^{3}: \; f'(0) = 0 \text{ but } f'(x) = 3x^{2} > 0 \text{ on both sides} \]
The derivative must change sign, not merely vanish. At the origin the cubic flattens momentarily and then carries on rising, so there is no turning point at all.
That distinction is the whole content of this section's central theorem, which runs one way only: an interior extremum forces a zero derivative, but a zero derivative forces nothing. Candidates still have to be tested.
Concept
On a closed interval a continuous function attains a largest and a smallest value. Each of them occurs either where the derivative is zero, or where the derivative fails to exist, or at an endpoint — and nowhere else.
absolute extremum — The largest or smallest value a function attains on a given set. It is a value of the function, attained at some input, and on a closed interval a continuous function is guaranteed to attain both.
\[ \text{extremum at } c \;\Longrightarrow\; f'(c) = 0, \text{ or } f'(c) \text{ undefined, or } c \text{ an endpoint} \]
That the list is complete is what makes the method work. Rather than examining infinitely many inputs, you evaluate the function at a handful of candidates and compare.
Figure (svg): The closed-interval method, showing the three places an absolute extremum can hide
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 317-320
Section
Section 1
Concept
A local maximum is a value larger than every value at nearby inputs. An absolute maximum is larger than every value on the whole domain. A local maximum need not be absolute, and an absolute one need not be local.
local and absolute extrema — A local maximum is largest within some interval around its input; an absolute maximum is largest on the entire domain under consideration. Endpoints can be absolute extrema without being local ones.
\[ \text{local at } c: \; f(c) \ge f(x) \text{ for } x \text{ near } c \]
The endpoints are the reason the two notions come apart. A value at an endpoint has neighbours on only one side, so it can be the largest on the interval while being nothing special locally.
Figure (svg): A curve on a closed interval with local and absolute extrema marked separately
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 317-320 — absolute and local extrema
Picture it
A cubic on a closed interval.
Figure (svg): A curve on a closed interval with local and absolute extrema marked separately
The local maximum at x equal to 1 has value 4, and the right endpoint has value 20. Searching only for local extrema would have found the smaller of the two and missed the answer entirely.
Worked example
Example 4.13. Local first, then compare values.
\[ \text{For } f(x)=x^{3}-6x^{2}+9x \text{ on } [0,5], \text{ identify all local and absolute extrema.} \]
Differentiate
Why: Termwise.
\[ f'(x) = 3 x ^{2} - 12 x + 9 \]
Factor and solve
Why: Take out 3.
\[ 3(x - 1) (x - 3) = 0,\text{ so } x = 1, 3 \]
Evaluate at those inputs
Why: The candidates.
\[ f(1) = 4, f(3) = 0 \]
Evaluate at the endpoints
Why: Both ends of the interval.
\[ f(0) = 0, f(5) = 20 \]
Compare all four values
Why: Largest and smallest.
\[ \max 20\text{ at } x = 5; \min 0\text{ at } x = 0\text{ and } x = 3 \]
Figure (svg): A function whose absolute extremum is at an endpoint, missed by looking only at critical points
\[ \max = 20 \text{ at } x=5; \quad \min = 0 \text{ at } x = 0, 3 \]
Verify: check that the local extrema are distinguished from the absolute ones
Why: There is a local maximum at x equal to 1 with value 4, which is not the absolute maximum — the endpoint value of 20 is far larger. There is a local minimum at x equal to 3 with value 0, which IS an absolute minimum, but so is the endpoint value at 0. Two different inputs can attain the same absolute minimum, and both should be reported. Notice that the absolute maximum occurred where the derivative is 24, not zero, which is exactly why endpoints are checked separately.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 319-320
Sorting
For the cubic on the closed interval from 0 to 5.
Sort into buckets
Sort each marked point.
The endpoints are the interesting bucket. They can be absolute extrema without any derivative condition holding at them, which is precisely why the closed-interval method checks them as a separate category.
Worked example
Checkpoint 4.13. Change the interval and the answer changes.
\[ \text{Does } f(x)=x \text{ have absolute extrema on } (0,2)? \]
Examine the behaviour
Why: The function increases throughout.
Look for a largest value
Why: Values approach 2 but never reach it.
Look for a smallest value
Why: Values approach 0 but never reach it.
Diagnose
Why: The endpoints are excluded.
Figure (svg): The solution to Worked example a function with no absolute extremum shown as a ladder of expressions, one row per legal move
\[ \text{no absolute extrema on } (0,2) \]
Verify: check what closing the interval does
Why: On the closed interval from 0 to 2 the same function attains a minimum of 0 and a maximum of 2, both at endpoints. Nothing about the function changed; only whether the endpoints belong to the domain. This is the sharpest illustration of why the Extreme Value Theorem insists on a closed interval, and it shows that the failure is about the DOMAIN rather than about any misbehaviour of the function.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 320-320
Trap
\[ f(x)=x^{3}-6x^{2}+9x \text{ on } [0,5] \]
Report the absolute maximum as 5
Why: The student gives the input where it occurs.
\[ \text{maximum} = 5 \quad \text{(wrong)} \]
Five is where the maximum occurs; the maximum itself is the value there, which is 20.
\[ \text{the absolute maximum is } 20, \text{ attained at } x = 5 \]
State the VALUE, and say where it is attained
Why: An extremum is an output, not an input.
Both pieces of information are usually wanted, so the safe form names both. The distinction matters practically: a manufacturer wants to know the maximum profit AND the production level achieving it, and confusing them makes the answer useless.
Fill the middle
The cubic's largest value on the interval from 0 to 5.
Fill in the blanks
\text20 ___, \text___ x = 5
Why: The maximum is 20, the value the function attains; 5 is where it attains it. Reporting the input as the maximum is the standard slip and makes the answer say the wrong thing.
Two truths and a lie
All three are about the two kinds of extremum.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The cubic has a local maximum of 4 at x equal to 1, while its absolute maximum on the interval is 20. A local extremum only compares against nearby values, so it says nothing about what happens elsewhere on the interval — which is why an absolute extremum must be found by comparing values rather than by finding turning points.
Prediction
Commit before reasoning.
Predict first
Why can a function attain its largest value at an endpoint even when the derivative there is not zero?
Correct: Because an endpoint has neighbours on one side only.
\[ f'(5) = 24 > 0 \text{ and yet } f(5) \text{ is the maximum on } [0,5] \]
Why: Fermat's theorem applies to INTERIOR extrema, where the function can be compared on both sides. At a right endpoint the function may be climbing steadily and simply run out of interval, so the largest value occurs there with a positive derivative. This is why the theorem's hypothesis says interior, and why the closed-interval method treats endpoints as a separate category rather than expecting them to be critical points.
Section
Section 2
Concept
A function continuous on a closed bounded interval attains an absolute maximum and an absolute minimum somewhere on it. Both hypotheses are needed: dropping either allows the extrema to fail to exist.
the Extreme Value Theorem — If f is continuous on a closed bounded interval, then there are inputs in that interval at which f attains an absolute maximum and an absolute minimum.
\[ f \text{ continuous on } [a,b] \;\Longrightarrow\; \exists c, d \in [a,b]: f(c) \le f(x) \le f(d) \]
Like the Intermediate Value Theorem of Section 2.4, this is an existence theorem. It promises that the extrema are there and says nothing about where, which is what the rest of the section supplies.
Figure (svg): The Extreme Value Theorem's two hypotheses, each shown failing
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 320-322 — the Extreme Value Theorem
Picture it
The theorem, with each hypothesis dropped in turn.
Figure (svg): The Extreme Value Theorem's two hypotheses, each shown failing
The middle curve has a jump and never reaches its supremum; the right one runs out of interval before attaining its. Neither failure is exotic, which is why the hypotheses are stated so carefully.
Worked example
Example 4.14. Change one condition at a time.
\[ \text{Does } f(x)=x \text{ attain extrema on } [0,2], \text{ on } (0,2), \text{ and does } 1/x \text{ on } [-1,1]? \]
Check the first case
Why: Continuous on a closed bounded interval.
Conclude for it
Why: The theorem applies.
\[ \min 0, \max 2,\text{ both at endpoints} \]
Check the second case
Why: Same function, open interval.
Conclude
Why: Values approach 0 and 2 without attaining them.
Check the third case
Why: Closed interval, but an asymptote at 0.
Conclude
Why: The function is unbounded.
Figure (svg): The solution to Worked example both hypotheses tested shown as a ladder of expressions, one row per legal move
\[ [0,2]: \text{ both}; \quad (0,2): \text{ neither}; \quad 1/x \text{ on } [-1,1]: \text{ neither} \]
Verify: confirm each failure traces to the dropped hypothesis
Why: The open interval fails because 2 is not in the domain, so the supremum is never attained even though the function is perfectly continuous. The reciprocal fails because it is unbounded near zero, so there is no largest value at all — and the interval is closed and bounded. Each counterexample isolates one hypothesis, which is what shows both are genuinely required rather than merely convenient.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 321-322
Sorting
Check continuity and closedness.
Sort into buckets
Sort each situation.
The last case fails on boundedness: the interval is closed but runs to infinity, and x squared has no largest value on it. Closed and bounded are two separate demands, and it is the pair that the theorem needs.
Worked example
Checkpoint 4.14. Existence without location.
\[ \text{A continuous } f \text{ on } [0,10] \text{ has } f(0)=3 \text{ and } f(10)=7. \text{ What follows about its maximum?} \]
Apply the theorem
Why: The hypotheses hold.
Ask where it is
Why: The theorem is silent.
Ask how large it is
Why: Also silent.
\[ \text{at least } 7,\text{ but possibly far more} \]
State what is known
Why: Existence and a lower bound only.
\[ \max\text{ exists and is at least } 7 \]
Figure (svg): The solution to Worked example what the theorem does not give shown as a ladder of expressions, one row per legal move
\[ \max f \ge 7, \text{ attained somewhere in } [0,10] \]
Verify: construct two functions consistent with the data
Why: A straight line from (0,3) to (10,7) has maximum 7 at the right endpoint. A function that rises to 100 in the middle and comes back down to 7 also fits the given values and has maximum 100 at an interior point. Both are continuous on the closed interval, so the theorem cannot distinguish them — which is exactly what it means for a theorem to guarantee existence without location. Finding the maximum needs the derivative, which is the next idea.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 322-322
Error analysis
A student reasons about a function on an open interval.
Annotate
On: \( f(x) = x \text{ is continuous on } (0,2), \text{ so it attains a maximum there} \)
Both hypotheses must be checked before the theorem is invoked. Continuity alone is not enough, and the open-interval failure is easy to miss because the function itself is perfectly well behaved.
Fill the middle
The function x on the open interval from 0 to 2.
Fill in the blanks
\textno \quad \text___ \; ___
Why: The interval is open, so the endpoints are excluded and the values approach 2 without attaining it. The theorem needs both hypotheses, and this one supplies only the first.
Two truths and a lie
All three are about the theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it reverses the theorem. The theorem says continuity GUARANTEES extrema; it does not say discontinuity prevents them. A function that jumps up to a single high value and drops back attains that maximum perfectly well despite being discontinuous. What is lost without continuity is the guarantee, not the possibility.
Prediction
Commit before reasoning.
Predict first
Why is an existence theorem useful if it does not locate the extrema?
Correct: Because it guarantees the search will succeed.
\[ \text{EVT: it exists} + \text{Fermat: only there} \;\Longrightarrow\; \text{compare and be done} \]
Why: Without the theorem, evaluating the function at every critical point and endpoint and picking the largest would prove nothing — the true maximum might be somewhere else, or might not exist. The theorem says it does exist and, combined with Fermat's theorem, that it must be among those candidates. That combination is what makes the closed-interval method complete rather than merely plausible.
Section
Section 3
Concept
If a function has a local extremum at an interior point and is differentiable there, its derivative at that point is zero. The converse is false: a zero derivative does not produce an extremum.
Fermat's theorem — If f has a local extremum at an interior point c and f prime of c exists, then f prime of c equals zero. The implication runs one way only.
\[ \text{local extremum at interior } c \text{ and } f'(c) \text{ exists} \;\Longrightarrow\; f'(c) = 0 \]
The argument is short. Just left of an interior maximum the function rises, so the derivative is at least zero there; just right it falls, so the derivative is at most zero. A derivative existing at the point must therefore be zero.
Figure (svg): An interior maximum with a horizontal tangent, and the argument that forces it
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 322-325 — Fermat's theorem
Picture it
An interior maximum with the slopes on either side.
Figure (svg): An interior maximum with a horizontal tangent, and the argument that forces it
The derivative is positive approaching from the left and negative approaching from the right. A derivative that exists at the peak cannot be both, so it is zero.
Worked example
Example 4.15. Solve for the candidates.
\[ \text{Find every interior point where } f(x)=x^{3}-6x^{2}+9x \text{ could have a local extremum.} \]
Note the function is differentiable everywhere
Why: It is a polynomial.
Apply Fermat's theorem
Why: An interior extremum needs a zero derivative.
\[ \text{solve } f'(x) = 0 \]
Differentiate and factor
Why: Take out 3.
\[ 3(x - 1) (x - 3) = 0 \]
Solve
Why: The two roots.
\[ x = 1\text{ and } x = 3 \]
State what has been established
Why: Candidates only.
Figure (svg): The solution to Worked example using the theorem to narrow the search shown as a ladder of expressions, one row per legal move
\[ f'(x) = 0 \text{ at } x = 1, 3 \]
Verify: confirm what the theorem does and does not settle
Why: The theorem has eliminated every other interior point in one step, which is an enormous saving — infinitely many inputs reduced to two. But it has not established that either candidate IS an extremum. Checking the graph shows both are, but the cubing function's origin is the standing counterexample where a candidate fails. The theorem narrows; it does not decide.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 323-324
Sorting
Check whether the derivative changes sign.
Sort into buckets
Sort each point where f' vanishes.
The pattern is the parity of the exponent: even powers turn and odd powers flatten. That is the same parity distinction that decided even and odd symmetry in Section 1.1, appearing again in a new setting.
Worked example
Checkpoint 4.15. A zero derivative with no extremum.
\[ \text{Show that } f(x)=x^{3} \text{ has } f'(0)=0 \text{ but no extremum at } 0. \]
Differentiate
Why: Power rule.
\[ f'(x) = 3 x ^{2} \]
Evaluate at the origin
Why: Zero.
\[ f'(0) = 0 \]
Examine the sign on both sides
Why: A square is non-negative.
\[ \text{f' } > 0\text{ for every } x\text{ other than } 0 \]
Conclude
Why: The function increases through the origin.
Check values nearby
Why: Negative to the left, positive to the right.
\[ f(-0.1) < 0 < f(0.1) \]
Figure (svg): The solution to Worked example the converse failing shown as a ladder of expressions, one row per legal move
\[ f'(0)=0 \text{ but } f'>0 \text{ on both sides} \]
Verify: state precisely what fails
Why: The value at the origin is 0, and the function takes both smaller and larger values arbitrarily nearby — so it is neither a local maximum nor a local minimum. What the vanishing derivative marks is a momentary flattening, and Section 4.5 will call it a point of inflection. The lesson is that Fermat's theorem is a necessary condition only, and every candidate must be tested by some further means.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 324-325
Trap
\[ f'(c) = 0 \]
Conclude there is an extremum at c
Why: The student reverses the implication.
\[ f(x) = x^{3}: \; f'(0)=0 \text{ and yet no extremum} \]
The cubing function flattens at the origin and carries on rising, so the vanishing derivative marks nothing.
\[ \text{extremum} \;\Longrightarrow\; f'(c)=0, \quad \text{but not conversely} \]
Treat a zero derivative as a candidate to be tested
Why: The theorem eliminates points; it does not confirm them.
The practical value is entirely in the elimination. Solving f prime equals zero reduces infinitely many inputs to a handful, and testing that handful is easy. Reading the implication backwards is the same error as reading the Intermediate Value Theorem as guaranteeing exactly one root.
Fill the middle
The cubic from the worked example, differentiated and factored.
Fill in the blanks
3(x-1)(x-3) = 0 \;\Longrightarrow\; x = 1 \text3 x = ___
Why: The two roots are the only interior candidates. Fermat's theorem has eliminated every other point in the interval, which is the entire practical value of the result.
Ranking
Why an interior maximum forces a zero derivative.
Put in order
Why: Step d is where the differentiability hypothesis is used: without it the two one-sided limits could differ, which is exactly what happens at a corner. The argument is a squeeze, and it is why the theorem needs the derivative to exist rather than merely the extremum to occur.
Prediction
Commit before reasoning.
Predict first
Why does Fermat's theorem require the extremum to be at an interior point?
Correct: Because the argument needs both sides, and an endpoint has only one.
\[ f'(5) = 24 \ne 0 \text{ and } f(5) \text{ is the maximum on } [0,5] \]
Why: The squeeze compares the left-hand and right-hand behaviour, and at an endpoint one of them lies outside the domain. So a function can climb steadily to a right endpoint and attain its maximum there with a positive derivative, as the cubic on the interval from 0 to 5 does. Endpoints are frequently extrema, which is exactly why the closed-interval method lists them as a third category rather than expecting Fermat's theorem to catch them.
Section
Section 4
Concept
A critical point is an interior input where the derivative is either zero or does not exist. Both kinds must be found, because an extremum can sit at a corner where no derivative exists at all.
critical point — An interior point of the domain at which the derivative is zero or fails to exist. Every interior local extremum occurs at a critical point, though not every critical point is an extremum.
\[ c \text{ critical} \iff f'(c) = 0 \text{ or } f'(c) \text{ undefined} \]
Searching only for zeros of the derivative is the commonest incomplete method. The absolute value has its minimum exactly where its derivative does not exist, and solving f prime equals zero finds nothing there.
Figure (svg): The two ways a critical point arises: a zero derivative and a derivative that fails to exist
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 323-326 — critical points
Picture it
A smooth turn, a corner, and a flattening.
Figure (svg): The two ways a critical point arises: a zero derivative and a derivative that fails to exist
All three are critical points and only the first two are extrema. The middle one would be missed by a method that only solves for zeros, and the third shows why candidates must still be tested.
Worked example
Example 4.16. Solving for zeros would find nothing.
\[ \text{Find the critical points of } f(x) = |x-1| + 0.4. \]
Rewrite as a piecewise function
Why: Split at the corner.
\[ x - 0.6\text{ for } x \ge 1; 1.4 - x\text{ for } x < 1 \]
Differentiate each piece
Why: Both linear.
Look for zeros
Why: Neither piece has a zero derivative.
\[ \text{no solutions of } f'(x) = 0 \]
Look for failures
Why: The one-sided derivatives disagree at the corner.
\[ f'(1)\text{ does not exist} \]
State the critical point
Why: The corner.
\[ x = 1 \]
Figure (svg): An absolute minimum at a corner, where the derivative does not exist
\[ x = 1: \; f' \text{ undefined} \]
Verify: confirm it is genuinely an extremum
Why: The function equals 0.4 at x equal to 1 and is larger everywhere else, since the absolute value is positive away from the corner. So this is an absolute minimum, attained at a point where the derivative does not exist. A method that solved only f prime equals zero would have found no candidates at all and concluded, wrongly, that there was no minimum.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 325-326
Sorting
Check for a zero or a failure, and that the point is in the domain.
Sort into buckets
Sort each candidate.
The reciprocal at 0 is the case to watch: the derivative certainly fails there, but 0 is not in the function's domain, so it is not a critical point. A critical point must be an input the function actually accepts.
Worked example
Checkpoint 4.16. Look for zeros and for failures.
\[ \text{Find the critical points of } f(x)=x^{2/3}(x-4). \]
Expand before differentiating
Why: Easier than the product rule here.
\[ f(x) = x ^{\frac{5}{3}} - 4 x ^{\frac{2}{3}} \]
Differentiate
Why: Power rule with rational exponents.
\[ f'(x) = (\frac{5}{3}) x ^{\frac{2}{3}} - (\frac{8}{3}) x ^{-\frac{1}{3}} \]
Combine over a common denominator
Why: Factor out the negative power.
\[ \frac{5 x - 8}{3 x ^{\frac{1}{3}}} \]
Find the zeros
Why: The numerator vanishes.
\[ x = \frac{8}{5} \]
Find the failures
Why: The denominator vanishes.
\[ x = 0 \]
Figure (svg): The solution to Worked example both kinds in one function shown as a ladder of expressions, one row per legal move
\[ x = \tfrac{8}{5} \text{ and } x = 0 \]
Verify: check that both are in the domain and classify them
Why: Both are in the domain, since x to the two thirds is defined for every real number. At x equal to 8 over 5 the derivative changes from negative to positive, giving a local minimum; at x equal to 0 the derivative is unbounded and the graph has a cusp, which turns out to be a local maximum. Missing the second candidate would have missed a genuine extremum, which is why the denominator must be examined as carefully as the numerator.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 326-326
Error analysis
A student finds the critical points of a function with a fractional power.
Annotate
On: \( f'(x) = \frac{5x-8}{3x^{1/3}} \;\Longrightarrow\; \text{critical point at } x = \tfrac85 \text{ only} \)
A quotient gives two conditions, not one. Setting the numerator to zero finds where the derivative vanishes; setting the denominator to zero finds where it fails, and both must be checked against the original function's domain.
Fill the middle
The derivative of a function with a fractional power, written as a quotient.
Fill in the blanks
f'(x) = \frac0___}: \; \text___ x=\tfrac85, \text___ x = ___
Why: The denominator vanishes at 0, so the derivative fails there — and since 0 is in the function's domain, it is a critical point. It turns out to be a local maximum, so missing it would lose a genuine extremum.
Two truths and a lie
All three are about critical points.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it describes only half the definition. Solving for zeros misses every corner and cusp, including the absolute value's minimum where no zero of the derivative exists at all. Both conditions must be checked, and for a derivative written as a quotient that means examining both the numerator and the denominator.
Prediction
Commit before reasoning.
Predict first
Why does the definition of a critical point include points where the derivative fails to exist?
Correct: Because an extremum can occur exactly at such a point.
\[ |x-1|: \; \text{minimum at } x=1 \text{ where } f' \text{ does not exist} \]
Why: The absolute value attains its minimum at its corner, where the one-sided derivatives are negative one and one and no derivative exists. Fermat's theorem does not apply there, since it requires differentiability — so the only way to catch such an extremum is to include failure points in the candidate list by definition. Section 3.2's three failure modes are exactly the shapes to look for: corners, cusps and vertical tangents.
Section
Section 5
Concept
To find the absolute extrema of a continuous function on a closed interval: find every critical point inside it, evaluate the function there and at both endpoints, and compare the values. The largest and smallest are the absolute extrema.
the closed-interval method — A complete procedure for absolute extrema: list the critical points in the interval and the two endpoints, evaluate f at each, and take the largest and smallest values.
\[ \max_{[a,b]} f = \max\{f(a), f(b), f(c_{1}), \ldots\} \]
No classification of the critical points is needed. Because the extrema must occur among the candidates, comparing values settles which is which without any first or second derivative test.
Figure (svg): The closed-interval method, showing the three places an absolute extremum can hide
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 326-328 — locating absolute extrema
Picture it
The complete list of places to check.
Figure (svg): The closed-interval method, showing the three places an absolute extremum can hide
The instruction at the bottom is the method's whole simplification: evaluate and compare. Deciding whether each critical point is a maximum, a minimum or neither is unnecessary work.
Worked example
Example 4.18. Candidates, values, comparison.
\[ \text{Find the absolute extrema of } f(x)=x^{2}-4x+3 \text{ on } [0,3]. \]
Differentiate and find the critical points
Why: A polynomial, so only zeros matter.
\[ f'(x) = 2 x - 4 = 0\text{ at } x = 2 \]
Check the critical point is in the interval
Why: Two lies between 0 and 3.
Evaluate at the critical point
Why: Four minus 8 plus 3.
\[ f(2) = -1 \]
Evaluate at both endpoints
Why: The two ends.
\[ f(0) = 3, f(3) = 0 \]
Compare the three values
Why: Largest and smallest.
\[ \max 3\text{ at } x = 0; \min - 1\text{ at } x = 2 \]
Figure (svg): The solution to Worked example the method in full shown as a ladder of expressions, one row per legal move
\[ \max = 3 \text{ at } x=0; \quad \min = -1 \text{ at } x=2 \]
Verify: check against the parabola's shape
Why: This parabola opens upward with its vertex at x equal to 2, so the minimum at the vertex is exactly what the shape demands. The maximum must then be at whichever endpoint is further from the vertex, and 0 is two units away while 3 is only one — so the left endpoint wins, with value 3 against 0. The geometry confirms the arithmetic completely, and notice that no derivative test was needed: comparing three numbers settled everything.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 327-328
Ranking
Finding absolute extrema on a closed interval.
Put in order
Why: Step a licenses the whole method by guaranteeing the extrema exist. Step c is easy to skip when the algebra produces roots automatically, and step e requires no classification of the critical points at all — comparing values is enough.
Worked example
Checkpoint 4.18. Discard what does not belong.
\[ \text{Find the absolute extrema of } f(x)=x^{3}-3x \text{ on } [0,3]. \]
Differentiate and solve
Why: Set to zero.
\[ 3 x ^{2} - 3 = 0,\text{ so } x = 1\text{ and } x = -1 \]
Discard candidates outside the interval
Why: Negative 1 is not in [0,3].
\[ \text{keep only } x = 1 \]
Evaluate at the surviving critical point
Why: One minus 3.
\[ f(1) = -2 \]
Evaluate at the endpoints
Why: Both ends.
\[ f(0) = 0, f(3) = 18 \]
Compare
Why: Three values.
\[ \max 18\text{ at } x = 3; \min - 2\text{ at } x = 1 \]
Figure (svg): The solution to Worked example a critical point outside the interval shown as a ladder of expressions, one row per legal move
\[ \max = 18 \text{ at } x=3; \quad \min = -2 \text{ at } x=1 \]
Verify: confirm discarding the other root was right
Why: At x equal to negative 1 the function equals 2, which is larger than the reported maximum candidate values other than 18 — but that input is not in the interval, so the function is not being considered there at all. Including it would answer a question about a different domain. Checking each critical point against the interval before evaluating is a necessary step, and it is easy to skip when the algebra produces the roots automatically.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 328-328
Trap
\[ f(x)=x^{3}-6x^{2}+9x \text{ on } [0,5]: \; f'(x)=0 \text{ at } x=1,3 \]
Compare only the critical values
Why: The student evaluates at 1 and 3 and stops.
\[ \max = 4 \text{ at } x=1 \quad \text{(wrong)} \]
The right endpoint gives f of 5 equal to 20, five times larger. The absolute maximum was never a critical point at all.
\[ \text{candidates: } x = 1, 3, 0, 5 \;\Longrightarrow\; \text{values } 4, 0, 0, 20 \]
Always include both endpoints in the candidate list
Why: Fermat's theorem covers only interior points.
On a closed interval the endpoints are genuine competitors, and for a function that is still climbing at the right end the maximum will always be there. Writing the candidate list as three categories — zeros, failures, endpoints — before evaluating anything makes the omission almost impossible.
Fill the middle
A quadratic on the closed interval from 0 to 3, with one critical point.
Fill in the blanks
\text3 x = 2 \text___, \; x = 0, \; x = ___
Why: Both endpoints belong on the list. Omitting them is the standard failure, and for a function still climbing at one end it loses the answer entirely.
Sorting
For a function on the closed interval from 0 to 3.
Sort into buckets
Sort each candidate.
The two discarded cases fail for different reasons and both are easy to get wrong: one is a genuine critical point in the wrong place, and the other is an ordinary interior point that no rule selects. Building the list from the three categories avoids both.
Prediction
Commit before reasoning.
Predict first
Why does the closed-interval method not require classifying each critical point?
Correct: Because the extrema must be among the candidates, so comparing values decides it.
\[ \text{EVT} + \text{Fermat} + \text{endpoints} \;\Longrightarrow\; \text{the list is complete} \]
Why: The Extreme Value Theorem guarantees the extrema exist and Fermat's theorem plus the endpoint rule guarantees they are among the listed candidates. So the largest value in the list IS the absolute maximum, whether or not that candidate is a local maximum. Classification matters when local extrema are wanted, which is Section 4.5's business, but for absolute extrema on a closed interval it is redundant work.
Comparison
Fill the blanks. Each supplies one piece of the method.
Comparison matrix
| Result | What it gives | What it does not |
|---|---|---|
| Extreme Value Theorem | the extrema exist | where they are |
| Fermat's theorem | interior extrema have zero derivative | that a zero derivative gives an extremum |
| The endpoint rule | endpoints are eligible too | any derivative condition there |
| The three together | a complete finite candidate list | nothing: comparing values finishes it |
No one of the three suffices. The theorem says the search will succeed, Fermat narrows it to finitely many places, and the endpoint rule closes the remaining gap.
Pattern
Given a continuous function on a closed interval and asked for its absolute extrema.
Steps two and three are two separate searches and both are required. Step four's endpoint addition is the one most often forgotten, and it costs the answer whenever the function is still climbing at an end.
Stewart, Calculus: Early Transcendentals 8e, §4.1 Maximum and Minimum Values §4.1, pp. 276-286
Check
Local against absolute.
Check your understanding
For f(x) = x^3 - 6x^2 + 9x on [0,5], what is the absolute maximum?
Answer: A
Why: The candidates give values 4, 0, 0 and 20; the endpoint value 20 is largest.
Check
Critical points. Two conditions, not one.
Check your understanding
Where are the critical points of f(x) = |x - 1|?
Answer: A
Why: The one-sided derivatives are -1 and 1 at the corner, so f'(1) does not exist.
Check
The theorem's hypotheses.
Check your understanding
Why does f(x) = x have no maximum on the open interval (0, 2)?
Answer: A
Why: Values approach 2 without attaining it, because 2 is not in the domain.
Real world
A delivery van's fuel use in litres per hundred kilometres depends on its speed: twenty-five, less six tenths per unit of speed, plus five thousandths times the speed squared. Legal speeds on the route run from 40 to 120 kilometres per hour.
Discussion prompt
Find the most and least efficient speeds in that range, and say what changes if the road's limit is lowered to 50 km/h.
Hint: This is the closed-interval method with a physical interval.
Answer:
\[ C'(v) = 0.01v - 0.6 = 0 \;\Longrightarrow\; v = 60 \]
The single critical point is at 60 km/h, which lies inside the range. Evaluating the three candidates:
\[ C(40) = 9, \qquad C(60) = 7, \qquad C(120) = 25 \]
So the most efficient speed is 60 km/h at 7 litres per hundred kilometres, and the least efficient is 120 km/h at 25 — nearly four times the consumption.
If the limit drops to 50 km/h, the interval becomes 40 to 50 and the critical point at 60 falls outside it. Discarding it leaves only the endpoints, and C(40) is 9 against C(50) of 7.5 — so the best available speed becomes 50, the upper endpoint, even though it is not a critical point at all.
That shift is the practical content of the endpoint rule. Constraining a problem does not merely trim the answer; it can move the optimum to a boundary where no derivative condition holds. Section 4.7 will meet this constantly, since almost every real optimisation problem carries constraints of exactly this kind.
Commit first
Answer, then rate your confidence honestly.
Predict first
A function has f'(c) = 0 at an interior point c. What follows?
Correct: It is a critical point, and may or may not be an extremum.
\[ x^{3}: \; f'(0)=0, \text{ no extremum}; \qquad x^{2}: \; f'(0)=0, \text{ a minimum} \]
Why: Fermat's theorem runs one way: an interior extremum forces a zero derivative, but a zero derivative forces nothing. The cubing function has a zero derivative at the origin and no extremum there, because the derivative does not change sign. So a zero derivative marks a candidate to be tested, not a conclusion. A constant function would need the derivative to vanish on a whole interval, not at a single point.
Explain it
They solved f prime equals zero, compared the two values, and reported the larger as the maximum — and got it wrong.
Discussion prompt
In four sentences or fewer, explain what they missed.
Hint: Ask them where the function is at the ends of the interval.
Answer:
Ask them what the function equals at the right-hand end of the interval. For the cubic on 0 to 5 it is 20, while their largest critical value was 4 — so the maximum is at a place where the derivative is 24, nowhere near zero.
Fermat's theorem only covers INTERIOR points, because its argument compares the function on both sides. An endpoint has neighbours on one side only, so a function can still be climbing when the interval runs out — which means endpoints must always be added to the candidate list by hand.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For endpoints, write the candidate list in three categories before evaluating anything. For failure points, examine a quotient derivative's denominator as carefully as its numerator. For local versus absolute, remember that absolute is decided by comparing values across the whole interval. For the theorem, check continuity and closed-and-bounded as two separate conditions. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw the cubic x cubed minus 6x squared plus 9x on the interval from 0 to 5, marking the two critical points and both endpoints with their values, and label which are local extrema and which are absolute. Write beside it the value of the derivative at the right endpoint and one sentence on why that does not prevent it being the maximum. Below, draw the three Extreme Value Theorem pictures — the successful case, a discontinuous one and an open interval — and write which hypothesis each violates. In the middle, draw an interior maximum with the slope arrows on either side and write out Fermat's argument in three lines. Beside it draw the cubing function at the origin and write why it is a counterexample to the converse. In the lower half, draw the absolute value's corner and write why solving f prime equals zero finds nothing there. At the bottom, write the closed-interval method as five numbered steps, with the three categories of candidate boxed. In a margin, write what an extremum IS — a value, not an input.
If your candidate list for the cubic has only two entries, the endpoints were forgotten — and the answer would be 4 instead of 20, which is the error this whole section exists to prevent.
Recap
Five things, and the last is a complete method rather than a technique.
| If you see | Then |
|---|---|
| A closed bounded interval and continuity | Both extrema exist |
| An open interval | The theorem does not apply |
| An interior extremum | The derivative there is zero or undefined |
| f'(c) = 0 | A candidate, not a conclusion |
| A derivative written as a quotient | Check the numerator AND the denominator |
| A corner | A critical point with no zero derivative |
| An absolute extremum question | Include both endpoints in the list |
Section 4.4 supplies the theorem that connects a function's average behaviour to its instantaneous behaviour. The Mean Value Theorem is proved from Fermat's theorem, and it is what finally justifies the reasoning of Section 4.5 that a positive derivative means an increasing function.
OpenStax Calculus Volume 1, §4.3 Maxima and Minima §4.3, pp. 317-328 — everything on these slides traces back here
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