Mean Value Theorem: Statement, Meaning and Worked Examples

Mean Value Theorem (MVT). If \( f \) is

1. continuous on the closed interval \( [a, b] \), and
2. differentiable on the open interval \( (a, b) \),

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

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

What it means

The right side is the slope of the secant line from \( (a, f(a)) \) to \( (b, f(b)) \): the average rate of change. The left side is the slope of the tangent at \( c \): the instantaneous rate. The theorem says somewhere in between, the curve is exactly parallel to the secant.

Speeding ticket version: if you drive 120 km in 1 hour, then at some instant your speedometer read exactly 120 km/h. The MVT guarantees it.

Why it works: slide the secant line

Draw the secant line through the two endpoints. Now slide a copy of it up or down, keeping it parallel. As you push it away from the curve, there is a last moment when it still touches the graph. At that point of contact the line just grazes the curve, so it is a tangent line, and it has the same slope as the secant. The \( x \)-coordinate of that touching point is a value of \( c \).

This picture also shows why the hypotheses are needed. A jump in the graph (not continuous) or a sharp corner (not differentiable) lets the curve “turn” without ever having a tangent with the right slope.

How to find c

1. Check the hypotheses (continuity and differentiability).
2. Compute the average rate \( \frac{f(b) - f(a)}{b - a} \).
3. Solve \( f'(c) = \) that value, and keep only solutions strictly inside \( (a, b) \).

You can check the derivative in step 3 with a derivative calculator.

Example 1

\( f(x) = x^3 - x \) on \( [0, 2] \). Polynomials are continuous and differentiable everywhere.

$$\frac{f(2) - f(0)}{2 - 0} = \frac{6 - 0}{2} = 3$$

Solve \( f'(c) = 3c^2 - 1 = 3 \): \( c^2 = \frac43 \), so \( c = \frac{2}{\sqrt3} \approx 1.1547 \), which lies in \( (0, 2) \). (The negative root is outside the interval.)

Example 2

\( f(x) = x^2 \) on \( [1, 4] \): average rate \( \frac{16 - 1}{3} = 5 \). \( 2c = 5 \) gives \( c = 2.5 \), exactly the midpoint. That’s a special property of parabolas.

Example 3: a logarithm

\( f(x) = \ln x \) on \( [1, e] \). The natural log is continuous and differentiable for \( x > 0 \), so both hypotheses hold. The average rate is

$$\frac{\ln e - \ln 1}{e - 1} = \frac{1}{e - 1}$$

Since \( f'(c) = \frac1c \), set \( \frac1c = \frac{1}{e-1} \) to get \( c = e - 1 \approx 1.718 \), which is between 1 and \( e \approx 2.718 \). Notice that \( c \) is not the midpoint here; the MVT only promises that \( c \) exists, not where it is.

When the MVT does not apply

\( f(x) = |x| \) on \( [-1, 1] \): the average rate is \( \frac{1 - 1}{2} = 0 \), but \( f'(x) \) is only ever \( \pm1 \). No \( c \) works, because \( |x| \) is not differentiable at 0. The hypotheses matter.

A second failure: \( f(x) = \frac1x \) on \( [-1, 1] \). The average rate is \( \frac{1 - (-1)}{2} = 1 \), but \( f'(x) = -\frac{1}{x^2} \) is always negative. The theorem fails because \( \frac1x \) is not even continuous at 0.

Rolle’s theorem: the special case

If additionally \( f(a) = f(b) \), the average rate is 0, so there is a \( c \) with \( f'(c) = 0 \): a horizontal tangent. That’s Rolle’s theorem, and it is usually proved first and then used to prove the MVT.

Proof of the Mean Value Theorem

Step 1: prove Rolle’s theorem. A continuous function on a closed interval reaches a maximum and a minimum there (the Extreme Value Theorem). If both happen at the endpoints, then since \( f(a) = f(b) \) the function never rises above or falls below that value, so it is constant and \( f' = 0 \) everywhere inside. Otherwise, a maximum or minimum occurs at some interior point \( c \). At an interior extreme point of a differentiable function the derivative must be zero, which is the same fact that makes critical points the candidates for extrema. So \( f'(c) = 0 \).

Step 2: tilt the picture. Let \( m \) be the secant slope and subtract the secant line from \( f \):

$$g(x) = f(x) - f(a) - m(x - a)$$

This \( g \) measures the vertical gap between the curve and the secant. It is continuous and differentiable wherever \( f \) is, and the gap is zero at both ends: \( g(a) = 0 \) and \( g(b) = f(b) - f(a) - m(b - a) = 0 \).

Step 3: apply Rolle. Rolle’s theorem gives a \( c \) in \( (a, b) \) with \( g'(c) = 0 \). But \( g'(x) = f'(x) - m \), so \( f'(c) = m \). That is the Mean Value Theorem.

Example 4: bounding a function

Suppose \( f(0) = 3 \) and \( f'(x) \le 5 \) for all \( x \). How large can \( f(2) \) be? By the MVT, \( \frac{f(2) - f(0)}{2 - 0} = f'(c) \) for some \( c \), so

$$f(2) = 3 + 2f'(c) \le 3 + 2(5) = 13$$

This type of question is common on exams. The MVT turns a bound on the derivative into a bound on the function.

Example 5: exactly one root

Show that \( x^3 + x - 1 = 0 \) has exactly one real solution. Let \( f(x) = x^3 + x - 1 \). Since \( f(0) = -1 \) and \( f(1) = 1 \), the Intermediate Value Theorem gives at least one root. If there were two roots, \( f \) would take equal values at two points, and Rolle’s theorem would give a \( c \) with \( f'(c) = 0 \). But \( f'(x) = 3x^2 + 1 \ge 1 \) is never zero. So there is exactly one root.

Example 6: proving an inequality

For any numbers \( a \) and \( b \), the MVT applied to \( \sin x \) gives \( \sin b - \sin a = \cos c\,(b - a) \) for some \( c \). Because \( |\cos c| \le 1 \),

$$|\sin b - \sin a| \le |b - a|$$

The sine function never changes faster than its input does.

Common mistakes

• Skipping the hypotheses. Always state that \( f \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \); graders look for it.
• Keeping a c outside the interval. Solving \( f'(c) = m \) often gives extra roots. Discard any not strictly between \( a \) and \( b \).
• Confusing the MVT with the IVT. The Intermediate Value Theorem is about function values; the MVT is about slopes.
• Flipping the fraction. The average rate is change in output over change in input, \( \frac{f(b) - f(a)}{b - a} \), not the reverse.
• Assuming c is the midpoint. That only happens for quadratics.

Why the MVT is important

It is the bridge between derivatives and function behavior:

• If \( f' > 0 \) on an interval, \( f \) is increasing there.
• If \( f' = 0 \) everywhere on an interval, \( f \) is constant, which is why antiderivatives differ only by \( + C \).
• It underpins error bounds for linear approximation and Taylor polynomials.
• It is a key step in the proof of the Fundamental Theorem of Calculus.

Mean Value Theorem calculator

For a larger version with a graph of the secant and tangent lines, use the Mean Value Theorem calculator.

Practice problems

Try these before checking the answers.

1. \( f(x) = x^2 + 2x \text{ on } [0, 4] \)
2. \( f(x) = \sqrt{x} \text{ on } [1, 9] \)
3. \( f(x) = x^3 \text{ on } [0, 3] \)
4. \( f(x) = \frac1x \) on \( [1, 3] \)
5. If \( f(1) = 10 \) and \( f'(x) \ge 2 \) on \( [1, 4] \), what is the smallest possible value of \( f(4) \)?
6. Find the value of \( c \) guaranteed by Rolle’s theorem for \( f(x) = x^2 - 4x \) on \( [0, 4] \).

Answers: (1) \( c = 2 \); (2) \( c = 4 \); (3) \( c = \sqrt3 \); (4) \( c = \sqrt3 \); (5) \( 16 \); (6) \( c = 2 \).

FAQ

Can there be more than one c?

Yes. The theorem guarantees at least one. For example, \( \sin x \) on \( [0, 2\pi] \) has average rate 0 and two values, \( c = \frac\pi2 \) and \( \frac{3\pi}{2} \).

What’s the Mean Value Theorem for integrals?

For continuous \( f \), there’s a \( c \) where \( f(c) \) equals the average value \( \frac{1}{b-a}\int_a^b f(x)\,dx \).

What is the difference between Rolle’s theorem and the Mean Value Theorem?

Rolle’s theorem is the special case where \( f(a) = f(b) \), so the guaranteed tangent is horizontal. The MVT allows any endpoint values and guarantees a tangent parallel to the secant line.

Why does differentiability only need to hold on the open interval?

The argument only uses derivatives at interior points, where \( c \) lives. That is why \( \sqrt x \) on \( [0, 4] \) still satisfies the MVT even though its derivative does not exist at 0.

How is the MVT different from the Intermediate Value Theorem?

The Intermediate Value Theorem says a continuous function hits every value between \( f(a) \) and \( f(b) \). The MVT says the derivative hits the average slope. One is about heights, the other about slopes.

Further reading

• The Mean Value Theorem (Paul’s Online Math Notes) — Rolle’s theorem, the MVT and several worked applications
• Average Function Value (Paul’s Online Math Notes) — the integral version of the Mean Value Theorem with examples