Mathematics guide, Unit 5, step 17 of 33Previous step Next step

Step 17: The Mean Value Theorem and Extreme Values

By the EconTutorials team. 12-minute read with 6 practice questions.

SyllabusAP 5.1–5.5 · IB SL 5.2

In 30 seconds

The Mean Value Theorem says that if $ f $ is continuous on $ [a, b] $ and differentiable on $ (a, b) $, then somewhere in between the tangent is parallel to the chord: $ f'(c) = \dfrac{f(b)-f(a)}{b-a} $ for at least one $ c $.
The Extreme Value Theorem says a function that is continuous on a closed interval has an absolute maximum and an absolute minimum there. Inside the interval they can only occur at critical points, where $ f’ = 0 $ or $ f’ $ does not exist.
The sign of $ f’ $ tells you where $ f $ is increasing or decreasing. The first derivative test classifies a critical point by the sign change of $ f’ $: $ + $ to $-$ is a local maximum, $-$ to $ + $ a local minimum, no change means neither.
Exam point: on a closed interval, use the candidates test: evaluate $ f $ at every critical point and at both end points, then compare. The end points often win.

Video lesson coming soonA short animated lesson for this step will appear here.

The Mean Value Theorem

A lorry covers 180 km in 2 hours, an average of 90 km per hour. Must its speedometer have read exactly 90 at some moment? Yes. If it had gone faster than 90 the whole way it would have covered more than 180 km, and if it had always gone slower it would have covered less. Somewhere along the road, the instantaneous speed equalled the average speed. That common-sense argument is the Mean Value Theorem.

In the language of Step 11, the average rate of change over $ [a, b] $ is the gradient of the chord (often called the secant) joining the end points. The instantaneous rate of change at $ c $ is the gradient of the tangent there. The theorem says the two must agree somewhere.

Key termMean Value Theorem (MVT): if $ f $ is continuous on the closed interval $ [a, b] $ and differentiable on the open interval $ (a, b) $, then there is at least one number $ c $ with $ a < c < b $ such that $ f'(c) = \dfrac{f(b)-f(a)}{b-a} $.

The special case where $ f(a) = f(b) $ is called Rolle’s theorem: the chord is horizontal, so somewhere in between $ f'(c) = 0 $. If a smooth curve starts and finishes at the same height, it must level off at least once on the way.

Worked example

Show that $ f(x) = x^3-3x $ satisfies the conditions of the MVT on $ [0, 3] $, and find the value of $ c $ that it guarantees.

Step 1$ f $ is a polynomial, so it is continuous on $ [0, 3] $ and differentiable on $ (0, 3) $. The conditions hold.

Step 2Chord gradient: $ f(3) = 27-9 = 18 $ and $ f(0) = 0 $, so $ \dfrac{18-0}{3-0} = 6 $.

Step 3Solve $ f'(c) = 3c^2-3 = 6 $: $ c^2 = 3 $, so $ c = \pm\sqrt{3} $.

Step 4Only $ c = \sqrt{3} \approx 1.732 $ lies in $ (0, 3) $; reject $ -\sqrt{3} $.

At $ x = \sqrt{3} $ the tangent has gradient 6, the same as the chord. Since $ f(\sqrt{3}) = 3\sqrt{3}-3\sqrt{3} = 0 $, that tangent is $ y = 6(x-\sqrt{3}) $, as Figure 1 shows.

The MVT for y = x³ − 3x on [0, 3]

y = x³ − 3xchord from (0, 0) to (3, 18): gradient 6tangent at x = √3: gradient 6

Figure 1. The tangent at c = √3 is parallel to the chord joining the end points: both have gradient 6, the average rate of change over [0, 3].

The MVT is an existence theorem. It promises that a $ c $ exists without saying where, which is exactly what AP free-response questions exploit. You are given a table of values of a differentiable function and asked whether $ f'(c) $ must equal some number. Work out the chord gradient between two rows, check the conditions, and quote the theorem.

Common mistakeApplying the MVT without checking the conditions. For $ f(x) = |x-1| $ on $ [0, 3] $, the chord gradient is $ \dfrac{2-1}{3} = \tfrac{1}{3} $, but $ f'(x) $ is only ever $ -1 $ or $ +1 $. The theorem has not failed: $ f $ has a corner at $ x = 1 $, so it is not differentiable on $ (0, 3) $ and the MVT does not apply. Always write “$ f $ is continuous on $ [a, b] $ and differentiable on $ (a, b) $” before you use it.

The MVT also explains a fact used in the rest of this step. Suppose $ f'(x) > 0 $ everywhere on an interval, and take any $ x_1 < x_2 $ there. The MVT gives $ f(x_2)-f(x_1) = f'(c)(x_2-x_1) $ for some $ c $ between them. Both factors on the right are positive, so $ f(x_2) > f(x_1) $. A positive derivative really does mean an increasing function.

The Extreme Value Theorem and critical points

The MVT is about gradients. The next theorem is about heights: when can we be sure a function has a largest and a smallest value?

Key termExtreme Value Theorem (EVT): if $ f $ is continuous on the closed interval $ [a, b] $, then $ f $ has at least one absolute maximum value and at least one absolute minimum value on $ [a, b] $.

Both conditions matter. On the open interval $ (0, 2) $, $ f(x) = x $ gets as close as you like to 0 and to 2 without ever reaching them, so it has no maximum and no minimum. On $ [-1, 1] $, the function $ \tfrac{1}{x^2} $ (with, say, $ f(0) = 0 $) is not continuous at 0 and grows without bound there, so it has no maximum. Notice the word “absolute”: the absolute maximum is the largest value on the whole interval, while a local maximum only beats its near neighbours.

Where can an extreme value sit? Either at an end point, or inside the interval at a point where the graph turns. At an interior turning point the tangent is horizontal ($ f’ = 0 $), or the graph has a corner or cusp ($ f’ $ does not exist).

Key termCritical point: a value $ c $ in the domain of $ f $ where $ f'(c) = 0 $ or $ f'(c) $ does not exist. A stationary point is the special case $ f'(c) = 0 $.

Worked example

Find the critical points of $ f(x) = x^{1/3}(x-4) $.

Step 1Expand first: $ f(x) = x^{4/3}-4x^{1/3} $.

Step 2Differentiate: $ f'(x) = \tfrac{4}{3}x^{1/3}-\tfrac{4}{3}x^{-2/3} = \tfrac{4}{3}x^{-2/3}(x-1) = \dfrac{4(x-1)}{3x^{2/3}} $.

Step 3$ f'(x) = 0 $ when $ x = 1 $. $ f'(x) $ does not exist at $ x = 0 $, where the denominator is zero, but $ f(0) = 0 $ is defined.

There are two critical points, $ x = 0 $ and $ x = 1 $. Being critical only makes a point a candidate; the next section shows that just one of these is an extremum.

Increasing, decreasing and the first derivative test

Between critical points, $ f’ $ keeps one sign, because a continuous $ f’ $ can only change sign by passing through zero. So to find where $ f $ rises and falls, mark the critical points on a number line and test one value of $ f’ $ in each interval. This is a sign diagram.

\[ \begin{aligned} f'(x) &> 0 \text{ on an interval} \;\Rightarrow\; f \text{ is increasing there} \\ f'(x) &< 0 \text{ on an interval} \;\Rightarrow\; f \text{ is decreasing there} \end{aligned} \]

Key termFirst derivative test: at a critical point $ c $ of a continuous function, if $ f’ $ changes from positive to negative, $ f $ has a local maximum at $ c $; if $ f’ $ changes from negative to positive, a local minimum; if $ f’ $ does not change sign, neither.

Worked example

For $ f(x) = x^4-8x^2 + 3 $, find the intervals on which $ f $ is increasing and decreasing, and classify its stationary points.

Step 1$ f'(x) = 4x^3-16x = 4x(x-2)(x + 2) $, which is zero at $ x = -2, 0, 2 $.

Step 2Test a value in each interval: $ f'(-3) = -60 < 0 $, $ f'(-1) = 12 > 0 $, $ f'(1) = -12 < 0 $, $ f'(3) = 60 > 0 $.

Step 3So $ f $ is decreasing for $ x < -2 $ and $ 0 < x < 2 $, and increasing for $ -2 < x < 0 $ and $ x > 2 $.

Step 4At $ x = \pm 2 $, $ f’ $ goes from $-$ to $ + $: local minima, $ f(\pm 2) = 16-32 + 3 = -13 $. At $ x = 0 $, $ f’ $ goes from $ + $ to $-$: a local maximum, $ f(0) = 3 $.

The graph is a W shape with minima at $ (-2, -13) $ and $ (2, -13) $ and a local maximum at $ (0, 3) $. In Figure 2, each zero of $ f’ $ lines up with a turning point of $ f $.

f(x) = x⁴ − 8x² + 3 and its derivative

f(x) = x⁴ − 8x² + 3f′(x) = 4x³ − 16x

Figure 2. Where the dashed graph of f′ is below the axis, f is decreasing; where it is above, f is increasing. Each zero of f′ with a sign change is a turning point of f.

Now return to $ f(x) = x^{1/3}(x-4) $, with $ f'(x) = \dfrac{4(x-1)}{3x^{2/3}} $. The denominator is never negative, so the sign of $ f’ $ is the sign of $ x-1 $: negative for $ x < 0 $, still negative for $ 0 < x < 1 $, positive for $ x > 1 $. There is a local minimum at $ x = 1 $, where $ f(1) = -3 $. At $ x = 0 $ the sign does not change, so there is no extremum; the curve passes through the origin with a vertical tangent.

Common mistakeAssuming that $ f'(c) = 0 $ means a maximum or minimum. For $ f(x) = x^3 $, $ f'(0) = 0 $ but $ f'(x) = 3x^2 $ is positive on both sides, so the curve keeps rising through the origin. Without a sign change there is no turning point.

Exam tipAn AP justification must name the function and the sign change: “$ f $ has a relative minimum at $ x = 2 $ because $ f’ $ changes from negative to positive there.” “$ f'(2) = 0 $” on its own earns nothing, and neither does “the graph goes down then up”. Increasing and decreasing functions are also tested in Cambridge 9709 Paper 1 (1.7) and IB AA SL 5.2. The MVT and EVT by name are AP topics; IB and Cambridge use the ideas without asking you to quote the theorems.

Absolute extrema: the candidates test

On a closed interval the EVT promises an absolute maximum and minimum, and the previous section shows they can only occur at critical points or end points. So there is no need to classify anything: list every candidate, evaluate $ f $ at each, and compare.

Step	Candidates test for a continuous $ f $ on $ [a, b] $
1	Find the critical points of $ f $ that lie inside $ (a, b) $. Ignore any outside the interval.
2	Evaluate $ f $ at each of them and at both end points $ a $ and $ b $.
3	The largest value is the absolute maximum; the smallest is the absolute minimum. State the value and where it occurs.

Worked example

Find the absolute maximum and minimum values of $ g(x) = 2x^3-9x^2 + 12x $ on $ [0, 3] $.

Step 1$ g'(x) = 6x^2-18x + 12 = 6(x-1)(x-2) $, so the critical points are $ x = 1 $ and $ x = 2 $, both inside $ (0, 3) $.

Step 2Candidates: $ g(0) = 0 $, $ g(1) = 2-9 + 12 = 5 $, $ g(2) = 16-36 + 24 = 4 $, $ g(3) = 54-81 + 36 = 9 $.

Step 3Largest: 9 at $ x = 3 $. Smallest: 0 at $ x = 0 $.

Both absolute extrema are at end points. The local maximum at $ (1, 5) $ is beaten by the right-hand end, as Figure 3 shows.

Candidates for g(x) = 2x³ − 9x² + 12x on [0, 3]

y = g(x) on [0, 3]

Figure 3. The four candidates. The interior turning points give local extrema only; the absolute maximum and minimum are both at end points of the interval.

An economics example: a capacity limit. A workshop’s profit is $ \pi(Q) = -Q^2 + 40Q-100 $ dollars a day for $ Q $ items, but its machines can make at most 15 items a day, so $ 0 \le Q \le 15 $. The only stationary point, $ \pi'(Q) = -2Q + 40 = 0 $ at $ Q = 20 $, is outside the feasible range. The candidates are the end points: $ \pi(0) = -100 $ and $ \pi(15) = -225 + 600-100 = 275 $. Profit is greatest at full capacity, $275 a day. Because $ \pi’ > 0 $ throughout, every extra item adds to profit, and the constraint, not the calculus, decides the answer.

Exam tipLay the candidates out in a short table of $ x $ and $ f(x) $. AP graders look for the candidates (including end points) and a clear conclusion such as “the absolute maximum value is 9, at $ x = 3 $”. Asked for the maximum value, give $ f $; asked where, give $ x $.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Find the value of $ c $ guaranteed by the Mean Value Theorem for $ f(x) = x^2 + 2x $ on $ [1, 4] $. Show worked answer

$ f $ is a polynomial, so the conditions hold.
Chord gradient: $ \dfrac{f(4)-f(1)}{4-1} = \dfrac{24-3}{3} = 7 $.
$ f'(c) = 2c + 2 = 7 $, so $ c = 2.5 $, which lies in $ (1, 4) $.

EasyQ2. Find the critical points of $ f(x) = x^3-12x + 1 $. Show worked answer

$ f'(x) = 3x^2-12 $, which exists for every $ x $.
$ 3x^2-12 = 0 $ gives $ x^2 = 4 $.
Critical points: $ x = -2 $ and $ x = 2 $.

MediumQ3. For $ f(x) = x^3-6x^2 + 9x + 2 $, find where $ f $ is increasing and decreasing, and use the first derivative test to classify its stationary points. Show worked answer

$ f'(x) = 3x^2-12x + 9 = 3(x-1)(x-3) $, zero at $ x = 1 $ and $ x = 3 $.
$ f'(0) = 9 > 0 $, $ f'(2) = -3 < 0 $, $ f'(4) = 9 > 0 $.
Increasing for $ x < 1 $ and $ x > 3 $; decreasing for $ 1 < x < 3 $.
At $ x = 1 $, $ f’ $ changes $ + $ to $-$: local maximum $ (1, 6) $.
At $ x = 3 $, $ f’ $ changes $-$ to $ + $: local minimum $ (3, 2) $.

MediumQ4. Find the absolute maximum and minimum values of $ h(x) = x^3-3x^2 + 1 $ on $ [-1, 4] $. Show worked answer

$ h'(x) = 3x^2-6x = 3x(x-2) $: critical points $ x = 0 $ and $ x = 2 $, both in the interval.
Candidates: $ h(-1) = -1-3 + 1 = -3 $, $ h(0) = 1 $, $ h(2) = 8-12 + 1 = -3 $, $ h(4) = 64-48 + 1 = 17 $.
Absolute maximum 17, at $ x = 4 $.
Absolute minimum $ -3 $, reached twice: at $ x = -1 $ and at $ x = 2 $.

MediumQ5. Show that $ f(x) = xe^{-x} $ has exactly one stationary point, and decide whether it is a maximum or a minimum. Show worked answer

Product rule: $ f'(x) = e^{-x}-xe^{-x} = (1-x)e^{-x} $.
$ e^{-x} > 0 $ always, so $ f'(x) = 0 $ only at $ x = 1 $: exactly one stationary point.
The sign of $ f’ $ is the sign of $ 1-x $: positive for $ x < 1 $, negative for $ x > 1 $.
$ f’ $ changes $ + $ to $-$, so $ \left(1, \tfrac{1}{e}\right) $ is a maximum, $ f(1) = e^{-1} \approx 0.368 $.

ExamQ6. The function $ g $ is differentiable for all real $ t $. Selected values are $ g(0) = 4 $, $ g(2) = 10 $, $ g(5) = 10 $, $ g(8) = 1 $. (a) Find the average rate of change of $ g $ over $ [0, 5] $. [1] (b) Must there be a value $ c $, $ 0 < c < 8 $, with $ g'(c) = -\tfrac{3}{8} $? Justify your answer. [2] (c) Explain why there must be a value $ r $, $ 2 < r < 5 $, with $ g'(r) = 0 $. [2] (d) Explain why $ g $ has an absolute maximum on $ [0, 8] $, and state a lower bound for its value. [2] Show worked answer

(a) $ \dfrac{g(5)-g(0)}{5-0} = \dfrac{10-4}{5} = 1.2 $. [B1]
(b) $ g $ is differentiable, so it is continuous on $ [0, 8] $ and differentiable on $ (0, 8) $. [M1]
$ \dfrac{g(8)-g(0)}{8-0} = \dfrac{1-4}{8} = -\tfrac{3}{8} $, so yes, by the Mean Value Theorem such a $ c $ exists. [A1]
(c) $ g $ is continuous on $ [2, 5] $ and differentiable on $ (2, 5) $. [M1]
$ \dfrac{g(5)-g(2)}{5-2} = 0 $, so by the MVT (Rolle’s theorem) some $ r $ in $ (2, 5) $ has $ g'(r) = 0 $. [A1]
(d) $ g $ is continuous on the closed interval $ [0, 8] $, so by the Extreme Value Theorem it has an absolute maximum there. [B1]
That maximum is at least the largest known value, so it is at least 10. [B1]
Total: 7 marks.

Check yourself

1. What value of $ c $ does the Mean Value Theorem give for $ f(x) = x^2 $ on $ [0, 4] $?

The chord gradient is $ \tfrac{16-0}{4} = 4 $, and $ f'(c) = 2c = 4 $ gives $ c = 2 $, the midpoint. For a quadratic the MVT point is always the midpoint of the interval.

2. Which condition guarantees that $ f $ has an absolute maximum on an interval?

The EVT needs continuity on a closed interval. $ f(x) = x $ is differentiable and continuous on $ (0, 1) $ but has no maximum there.

3. $ f'(x) = (x-1)^2(x + 2) $. Which statement is true?

$ (x-1)^2 \ge 0 $, so the sign of $ f’ $ is the sign of $ x + 2 $. It changes from $-$ to $ + $ at $ -2 $ (minimum) and does not change at 1.

Your score: 0 out of 3

The short version

MVT: if $ f $ is continuous on $ [a, b] $ and differentiable on $ (a, b) $, then $ f'(c) = \dfrac{f(b)-f(a)}{b-a} $ for some $ c $ in $ (a, b) $. Check and state the conditions.
EVT: continuous on a closed interval means an absolute maximum and an absolute minimum exist.
Critical points are where $ f’ = 0 $ or $ f’ $ does not exist. They are candidates, not guaranteed extrema.
$ f’ > 0 $: increasing; $ f’ < 0 $: decreasing. First derivative test: $ + $ to $-$ is a local maximum, $-$ to $ + $ a local minimum.
Candidates test on $ [a, b] $: compare $ f $ at the critical points inside the interval and at both end points.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step	Candidates test for a continuous \( f \) on \( [a, b] \)
1	Find the critical points of \( f \) that lie inside \( (a, b) \). Ignore any outside the interval.
2	Evaluate \( f \) at each of them and at both end points \( a \) and \( b \).
3	The largest value is the absolute maximum; the smallest is the absolute minimum. State the value and where it occurs.

Step 17: The Mean Value Theorem and Extreme Values

The Mean Value Theorem

The Extreme Value Theorem and critical points

Increasing, decreasing and the first derivative test

Absolute extrema: the candidates test

Practise: 6 questions

Check yourself

The short version

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