The Mean Value Theorem | Math Coach Amy

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Key Ideas Easy Medium Hard

\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)

What to Remember

State both hypotheses

The Mean Value Theorem needs \(f\) continuous on \([a, b]\) and differentiable on \((a, b)\).

Then there is at least one \(c\) in \((a, b)\) with

\(f'(c) = \dfrac{f(b) - f(a)}{b - a}\)

Rolle's Theorem is the flat case

If the hypotheses hold and \(f(a) = f(b)\), then \(f'(c) = 0\) for some \(c\) in \((a, b)\).

Find c from the derivative equation

Compute the average rate first. Set \(f'(x)\) equal to that number. Keep only the solutions that sit in the open interval.

A failed hypothesis blocks the guarantee

A corner, a hole, or a vertical asymptote inside the interval means the theorem does not apply. A derivative value might still occur, but the theorem does not promise it.

Level 1

Easy

The function is a polynomial, so both hypotheses hold. Find c.

Example

Find a value \(c\) guaranteed by the Mean Value Theorem for \(f(x) = x^2\) on \([1, 3]\).

\(f\) is a polynomial, so it is continuous on \([1, 3]\) and differentiable on \((1, 3)\).

\(\dfrac{f(3) - f(1)}{3 - 1} = \dfrac{9 - 1}{2}\)

\(4\)

\(f'(x) = 2x\)

\(2x = 4\)

\(c = 2\)

Answer: \(c = 2\)

Find \(c\) for \(f(x) = x^2\) on \([0, 2]\).

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A polynomial is continuous on the closed interval and differentiable on the open interval.

\(\dfrac{4 - 0}{2} = 2\)

\(2x = 2\)

\(c = 1\)

Answer: \(c = 1\)

Find \(c\) for \(f(x) = x^2 + x\) on \([0, 4]\).

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\(f(4) = 20\) and \(f(0) = 0\)

\(\dfrac{20}{4} = 5\)

\(f'(x) = 2x + 1\)

\(2x + 1 = 5\)

\(c = 2\)

Answer: \(c = 2\)

Find \(c\) for \(f(x) = \sqrt{x}\) on \([1, 4]\).

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\(\sqrt{x}\) is continuous on \([1, 4]\) and differentiable on \((1, 4)\).

\(\dfrac{2 - 1}{3} = \dfrac{1}{3}\)

\(f'(x) = \dfrac{1}{2\sqrt{x}}\)

\(\dfrac{1}{2\sqrt{x}} = \dfrac{1}{3}\)

\(2\sqrt{x} = 3\)

\(x = \dfrac{9}{4}\)

Answer: \(c = \dfrac{9}{4}\)

Does the Mean Value Theorem apply to \(f(x) = |x|\) on \([-1, 1]\)?

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\(f\) is continuous on \([-1, 1]\).

\(f\) has a corner at \(x = 0\).

\(f\) is not differentiable on \((-1, 1)\).

The Mean Value Theorem does not apply.

Answer: no

Does the Mean Value Theorem apply to \(f(x) = \dfrac{1}{x}\) on \([-1, 1]\)?

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\(f\) is undefined at \(x = 0\).

\(f\) is not continuous on \([-1, 1]\).

The Mean Value Theorem does not apply.

Answer: no

Level 2

Medium

A cubic, Rolle's Theorem, or a justification that stops at existence.

Example

Find the value \(c\) in \((0, 2)\) for \(f(x) = x^3 - 3x\) on \([0, 2]\).

A polynomial meets both hypotheses.

\(f(2) = 8 - 6 = 2\) and \(f(0) = 0\)

\(\dfrac{2 - 0}{2} = 1\)

\(f'(x) = 3x^2 - 3\)

\(3x^2 - 3 = 1\)

\(x^2 = \dfrac{4}{3}\)

The solution in \((0, 2)\) is \(c = \dfrac{2}{\sqrt{3}}\).

Answer: \(c = \dfrac{2}{\sqrt{3}}\)

Find \(c\) guaranteed by Rolle's Theorem for \(f(x) = x^2 - 4x\) on \([0, 4]\).

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\(f(0) = 0\) and \(f(4) = 0\)

The hypotheses of Rolle's Theorem hold.

\(f'(x) = 2x - 4\)

\(2x - 4 = 0\)

\(c = 2\)

Answer: \(c = 2\)

Find \(c\) for \(f(x) = \sin x\) on \([0, \pi]\).

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\(f(0) = f(\pi) = 0\)

Rolle's Theorem applies.

\(f'(x) = \cos x\)

\(\cos c = 0\)

\(c = \dfrac{\pi}{2}\)

Answer: \(c = \dfrac{\pi}{2}\)

\(f\) is continuous on \([1, 5]\), differentiable on \((1, 5)\), \(f(1) = 2\), and \(f(5) = 14\). Explain why \(f'(c) = 3\) for some \(c\) in \((1, 5)\).

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Both hypotheses of the Mean Value Theorem hold.

\(\dfrac{f(5) - f(1)}{5 - 1} = \dfrac{12}{4} = 3\)

There is a \(c\) in \((1, 5)\) with \(f'(c) = 3\).

Answer: yes, \(f'(c) = 3\)

The values \(f(1) = 2\) and \(f(5) = 14\) are the same, but \(f\) is not differentiable at \(x = 3\) inside \([1, 5]\). Does the Mean Value Theorem guarantee \(f'(c) = 3\)?

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Differentiability on the open interval fails.

The Mean Value Theorem does not apply, so it does not guarantee that slope.

Answer: no

Which conclusion follows when \(f\) is continuous on \([0, 4]\), differentiable on \((0, 4)\), \(f(0) = 1\), and \(f(4) = 1\)?

A. \(f'(c) = 0\) for some \(c\) in \((0, 4)\)

B. \(f'(c) = 1\) for some \(c\) in \((0, 4)\)

C. \(f(2) = 1\)

D. \(f\) is constant

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The endpoint values are equal, so Rolle's Theorem applies.

\(f'(c) = 0\) for some \(c\) in \((0, 4)\).

Answer: A. \(f'(c) = 0\) for some c in \((0, 4)\)

Level 3

Hard

A radical equation for c, a uniqueness argument, and data that cannot happen.

Example

Find \(c\) for \(f(x) = x^3 + x\) on \([1, 3]\).

A polynomial meets both hypotheses.

\(f(3) = 30\) and \(f(1) = 2\)

\(\dfrac{28}{2} = 14\)

\(f'(x) = 3x^2 + 1\)

\(3x^2 + 1 = 14\)

\(x^2 = \dfrac{13}{3}\)

\(c = \sqrt{\dfrac{13}{3}}\)

Answer: \(c = \sqrt{\dfrac{13}{3}}\)

Find \(c\) for \(f(x) = \ln x\) on \([1, e]\).

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\(\ln x\) is continuous on \([1, e]\) and differentiable on \((1, e)\).

\(\dfrac{\ln e - \ln 1}{e - 1} = \dfrac{1}{e - 1}\)

\(f'(x) = \dfrac{1}{x}\)

\(\dfrac{1}{c} = \dfrac{1}{e - 1}\)

\(c = e - 1\)

\(e - 1\) is in \((1, e)\) because \(2 < e\).

Answer: \(c = e - 1\)

Explain why \(x^3 + x - 1 = 0\) has exactly one real solution.

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Let \(f(x) = x^3 + x - 1\).

\(f(0) = -1\) and \(f(1) = 1\).

f is continuous, and \(-1 < 0 < 1\), so the Intermediate Value Theorem gives a root in \((0, 1)\).

\(f'(x) = 3x^2 + 1 > 0\) for every \(x\), so f is strictly increasing.

A strictly increasing function crosses 0 only once.

If there were two roots, Rolle's Theorem would force \(f'(c) = 0\) between them, and that never happens.

Answer: exactly one real root

Suppose \(f(2) = 5\), \(f(6) = 5\), and \(f'(x) > 0\) for every \(x\) in \((2, 6)\). Can \(f\) be continuous on \([2, 6]\) and differentiable on \((2, 6)\)?

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Those hypotheses plus equal endpoint values would give a \(c\) with \(f'(c) = 0\).

The given derivative is always positive, so it is never 0.

No function can meet all of these conditions at once.

Answer: no

For \(f(x) = x^3 - 3x\) on \([0, 2]\), the Mean Value Theorem produces \(c = \dfrac{2}{\sqrt{3}}\). Confirm that this \(c\) is inside \((0, 2)\).

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\(\dfrac{4}{3} < 4\)

\(\sqrt{\dfrac{4}{3}} < 2\)

\(\dfrac{2}{\sqrt{3}} > 0\)

So \(c\) is in \((0, 2)\).

Answer: yes, inside \((0, 2)\)

\(g\) is continuous on \([2, 6]\), differentiable on \((2, 6)\), \(g(2) = -1\), and \(g(6) = 7\). What exact slope must \(g'\) hit on \((2, 6)\)?

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\(\dfrac{g(6) - g(2)}{6 - 2} = \dfrac{8}{4}\)

\(2\)

The Mean Value Theorem says \(g'(c) = 2\) for some \(c\) in \((2, 6)\).

Answer: \(2\)

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