The Mean Value Theorem

The Mean Value Theorem says that if a function is continuous on [a, b] and differentiable on (a, b), then at some point strictly between a and b the instantaneous rate equals the average rate. On its own that is a statement about one point; its consequences are what get examined: on an interval, a function whose derivative is never zero has at most one root, a function whose derivative is bounded cannot grow too fast, and two functions with the same derivative differ by a constant.

Start practising Browse all 11 questions

In this topic

Taught: Not in the course
Questions: 11
Total marks: 28
Suggested time: 49 min
Difficulty: Introductory to Capstone
Show the 3 hardest

Key results

The Intermediate Value Theorem: $Y \text{ between } f(a) \text{ and } f(b) \implies f(c) = Y \text{ for some } c \in [a, b]$
When $a < b$ and $f$ is continuous on the closed interval $[a, b]$ .
Rolle’s theorem: $f(a) = f(b) \implies f'(c) = 0 \text{ for some } c \in (a, b)$
When $a < b$ , $f$ is continuous on $[a, b]$ and $f$ is differentiable on $(a, b)$ .
The Mean Value Theorem: $f'(c) = \frac{f(b) - f(a)}{b - a} \text{ for some } c \in (a, b)$
When $a < b$ , $f$ is continuous on $[a, b]$ and $f$ is differentiable on $(a, b)$ .
Counting zeros: $\begin{aligned} &f \text{ has } n \text{ distinct zeros} \\ &\implies f' \text{ has at least } n - 1 \text{ zeros} \end{aligned}$
When $f$ is differentiable on an interval containing the zeros, the zeros of $f'$ lying between those of $f$ ; so if $f'(x) \neq 0$ on an interval, $f$ has at most one zero there.
Equal derivatives: $f'(x) = g'(x) \text{ for all } x \in (a, b) \implies f(x) = g(x) + C \text{ on } (a, b)$
For some constant $C$ , on an interval; with $g = 0$ , a function whose derivative is 0 on an interval is constant there.

Read the theoryCLP-1 §2.13 The Mean Value Theorem

Worked example

Standard3 marks · 5 min

How many real roots does the function below have?

g(x) = 4x - \sin(2x) + 1

Strategy

Existence from a sign change and the IVT; uniqueness because the derivative never vanishes, so Rolle’s theorem forbids two roots.

Working

$g(-1) = -4 - \sin(-2) + 1 = -3 + \sin 2 < 0, \qquad g(0) = 1 > 0$
sin 2 ≈ 0.91, so g(−1) ≈ −2.09. g is continuous, as a sum of continuous functions, so this sign change and the IVT give at least one root in (−1, 0).
$g'(x) = 4 - 2\cos(2x) \geq 4 - 2 = 2 > 0$
The derivative is bounded away from zero.
g is differentiable everywhere. If g had two roots a < b, Rolle’s theorem would give a c between them with g′(c) = 0, which is impossible. So there is at most one root, hence exactly one.

Answer

\text{exactly one real root}

Check

The line 4x + 1, rising at slope 4, meets the wave sin(2x), whose slope is at most 2, only once, near x = −0.444.

Where this goes wrong

Guessing that the oscillating sine term creates several crossings. The slope of 4x dominates: the derivative never drops below 2, so the graph only ever rises.

On the paper

Check both hypotheses by name, continuous on $[a, b]$ and differentiable on $(a, b)$ , before using Rolle's theorem or the Mean Value Theorem, then say what it gives.

Leans on

The Intermediate Value Theorem, from Continuity

Your turn

Challenging3 marks · 6 min

How many real roots does the function below have?

p(x) = x^{4} - 4x + 1

Answer this question

Practice questions

5 of the 11, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
The Mean Value Theorem guarantees a point $c$ in $(-1, 1)$ with $f'(c) = \dfrac{f(1) - f(-1)}{2}$ for three of the functions below. For which one does it make no guarantee?
Strategy
Check the hypotheses: continuity on [−1, 1] and a derivative at every point of (−1, 1). One failure inside the interval voids the guarantee.
Working
x², sin x and eˣ are continuous and differentiable on the whole real line, so the MVT applies to each on [−1, 1].
$|x| \text{ is not differentiable at } 0 \in (-1, 1)$
The hypothesis fails, so the theorem is silent. Indeed the secant slope is (1 − 1)/2 = 0 and |x| never has slope 0.
Answer
$f(x) = |x|$
The answer is the absolute value function.
Check
No such c can exist for |x|: its secant slope over [−1, 1] is 0, and its slope is −1 or 1 wherever it has one.
Where this goes wrong
Assuming that a function which fails the hypotheses must also fail the conclusion, or that it must still satisfy it. When a hypothesis fails, the theorem says nothing either way; here the conclusion happens to fail too, but that is not automatic.
Leans on
Differentiability and corners, from The definition of the derivative
Try a similar question
Let $f(x) = |x - 2|$ . Which statement about $f$ at $x = 2$ is correct?
Answer this question
Introductory2 marks · 3 min
Find the value of $c$ in $(1, 4)$ whose existence the Mean Value Theorem guarantees for $f(x) = x^{2}$ on $[1, 4]$ .
Strategy
Compute the average rate of change over the interval, then find where the instantaneous rate equals it.
Working
$\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = 5$
Slope of the secant line.
$f'(c) = 2c = 5 \quad\Longrightarrow\quad c = \tfrac{5}{2}$
Which lies in (1, 4), as the theorem promises.
Answer
$c = \frac{5}{2}$
The answer is five halves.
Check
For any quadratic the MVT point is the midpoint of the interval, and (1 + 4)/2 = 2.5.
Where this goes wrong
Setting f(c), not f′(c), equal to the secant slope 5, which gives √5. The theorem matches the slope of the tangent at c with the slope of the secant, so it is f′(c) that equals 5.
Leans on
Slope of a secant line, from The definition of the derivative
Try a similar question
Find the point $c$ promised by the Mean Value Theorem for $f(x) = \log x$ on the interval $[1, e]$ .
Answer this question
Standard2 marks · 3 min
Find the point $c$ promised by the Mean Value Theorem for $f(x) = \log x$ on the interval $[1, e]$ .
Strategy
Compute the secant slope using the exact values of log at the endpoints, then solve 1/c equals that slope.
Working
$\frac{f(e) - f(1)}{e - 1} = \frac{1 - 0}{e - 1} = \frac{1}{e - 1}$
Secant slope.
$f'(c) = \frac{1}{c} = \frac{1}{e - 1} \quad\Longrightarrow\quad c = e - 1$
Solve. Check: 1 < e − 1 ≈ 1.72 < e ≈ 2.72, inside the interval.
Answer
$c = e - 1$
The answer is e minus 1, about 1.718.
Check
The tangent slope at 1.718 is 1/1.718 ≈ 0.582, the same as the chord from (1, 0) to (e, 1), which rises 1 over a run of e − 1.
Where this goes wrong
Writing f′(c) = log c instead of 1/c, or forgetting that log 1 = 0. The derivative of the logarithm is the reciprocal.
Leans on
The derivative of log x, from The chain rule and logarithms
Try a similar question
Find the value of $c$ in $(1, 4)$ whose existence the Mean Value Theorem guarantees for $f(x) = x^{2}$ on $[1, 4]$ .
Answer this question
Challenging4 marks · 7 min
Use the Mean Value Theorem to show that $\sqrt{1 + x} < 1 + \dfrac{x}{2}$ for every $x > 0$ .
Strategy
The MVT expresses the change in g over [0, x] as x times a derivative value at an interior point. Because the derivative decreases, that value is below its value at 0, which gives the inequality.
Working
$g(t) = \sqrt{1 + t}, \qquad g'(t) = \frac{1}{2\sqrt{1 + t}}$
g is continuous on [0, x] and differentiable on (0, x) for any x > 0.
$\sqrt{1 + x} - 1 = g(x) - g(0) = x\,g'(c) = \frac{x}{2\sqrt{1 + c}} \quad\text{for some } c \in (0, x)$
Mean Value Theorem.
$c > 0 \Rightarrow \sqrt{1 + c} > 1 \Rightarrow \frac{x}{2\sqrt{1 + c}} < \frac{x}{2}$
x is positive, so dividing by a larger positive number gives a smaller result.
$\sqrt{1 + x} < 1 + \frac{x}{2}$
Rearrange.
Answer
$\sqrt{1 + x} < 1 + \frac{x}{2} \quad (x > 0)$
The answer is the inequality holds for every positive x.
What earns the marks
Applies the MVT to g(t) = √(1 + t) on [0, x], obtaining √(1 + x) − 1 = x · g′(c) = x / (2√(1 + c)) for some c in (0, x).
Uses c > 0 to bound the derivative: 1/(2√(1 + c)) < 1/2.
Concludes √(1 + x) − 1 < x/2 and rearranges to the required inequality.
Check
Squaring gives a quick confirmation: (1 + x/2)² = 1 + x + x²/4, which exceeds 1 + x for every x > 0.
Where this goes wrong
Squaring both sides and checking the algebra. That verifies the inequality but does not “use the Mean Value Theorem”, which is what the question asks for and what the marks are attached to.
Leans on
Derivative of a square root, from The chain rule and logarithms
Try a similar question
Let $f(x) = x^{2} - 4\pi x + 3\sin x$ . Show that there is a real number $c$ with $f'(c) = 0$ .
Answer this question
Capstone5 marks · 9 min
For which values of the constant $k$ does $f(x) = x^{3} - 3x + k$ have exactly one real root? Enter the smallest positive integer $k$ with this property.
Strategy
The two critical values control everything. A cubic with a local max and a local min has three roots when they straddle the axis, two when one sits on it, and one when both are on the same side. Express the condition in k.
Working
$f'(x) = 3x^{2} - 3 = 3(x - 1)(x + 1)$
Critical points at −1 and 1; f increases, decreases, then increases.
$f(-1) = -1 + 3 + k = k + 2, \qquad f(1) = 1 - 3 + k = k - 2$
The local maximum value and the local minimum value.
By the IVT and monotonicity on each piece: three roots if k − 2 < 0 < k + 2, i.e. −2 < k < 2; a repeated root if k = ±2; exactly one root if both critical values have the same sign, i.e. k > 2 or k < −2.
$\text{exactly one root} \iff |k| > 2;\qquad \text{smallest positive integer: } k = 3$
k = 3 gives f(−1) = 5 and f(1) = 1, both positive, so the graph crosses the axis only once, to the left of −1.
Answer
$|k| > 2;\quad k = 3$
The answer is exactly one root when the absolute value of k exceeds 2; the smallest positive integer is 3.
Check
x³ − 3x + 3 has its only real root near x = −2.10, while x³ − 3x + 2 = (x − 1)²(x + 2) has two.
Where this goes wrong
Answering k = 2. At k = 2 the local minimum sits exactly on the axis, giving a double root at x = 1 and a second root at x = −2: two distinct roots, not one.
Try a similar question
How many real roots does the function below have? $p(x) = x^{4} - 4x + 1$
Answer this question

Browse all 11 questions

Needs
Continuity
The definition of continuity, classifying discontinuities, the constants that make a piecewise function continuous, and the Intermediate Value Theorem.
28 questions
Needs
Differentiation rules
Power, sum, product and quotient rules, tangent lines and higher-order derivatives.
23 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

What earns the marks

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Related topics

Continuity

Differentiation rules