The mean value theorem

The statement

Mean 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

The right side is the average rate of change: the slope of the secant line from (a, f(a)) to (b, f(b)). The left side is the slope of the tangent line at c.

The hypotheses, and why they matter

1. Continuous on [a, b]: no breaks, jumps or holes, including at the endpoints.
2. Differentiable on (a, b): no corners, cusps or vertical tangents between the endpoints.

Drop either one and the conclusion can fail. Example: f(x) = |x| − 1 on [−1, 1] is continuous and f(−1) = f(1) = 0, but f has a corner at x = 0. Its slope is −1 on the left and +1 on the right, so there's no c where f′(c) = 0, the slope of the secant line.

The theorem guarantees that c exists; it doesn't tell you where. You find c by solving an equation.

Rolle's theorem

Rolle's theorem is the special case where the endpoints have the same height. If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there's at least one c in (a, b) with

In words: if a smooth curve starts and ends at the same height, it has a horizontal tangent somewhere in between. Example: f(x) = x³ − 4x on [−2, 2] has f(−2) = f(2) = 0. Solving f′(x) = 3x² − 4 = 0 gives c = ±2/√3 ≈ ±1.155, both inside (−2, 2).

What it means geometrically

Draw the secant line joining the two endpoints. The MVT says that at some point c between them, the tangent line to the curve is parallel to that secant line. Rolle's theorem is the case where the secant line is horizontal.

In motion terms: if your average speed over a trip was 50 mph, at some instant your speedometer read exactly 50 mph. Example: a rock dropped from 100 feet has height s(t) = −16t² + 100 and hits the ground at t = 5/2 seconds. Its average velocity is (0 − 100) / (5/2) = −40 ft/s. Solving s′(t) = −32t = −40 gives t = 5/4 seconds, when its instantaneous velocity equals the average.

Example 1: find c for f(x) = √x on [0, 9]

1. Check the hypotheses. √x is continuous on [0, 9] and differentiable on (0, 9).
2. Average rate of change: (f(9) − f(0)) / (9 − 0) = (3 − 0) / 9 = 1/3.
3. Derivative: f′(x) = 1 / (2√x).
4. Solve 1 / (2√c) = 1/3: 2√c = 3, so √c = 3/2 and c = 9/4.
5. Check: 9/4 = 2.25 is inside (0, 9).

Example 2: find c for f(x) = x³ + 2x² − x on [−1, 2]

1. Check the hypotheses. A polynomial is continuous and differentiable everywhere.
2. Endpoint values: f(2) = 8 + 8 − 2 = 14; f(−1) = −1 + 2 + 1 = 2.
3. Average rate of change: (14 − 2) / (2 − (−1)) = 12/3 = 4.
4. Derivative: f′(x) = 3x² + 4x − 1.
5. Solve 3c² + 4c − 1 = 4, so 3c² + 4c − 5 = 0. By the quadratic formula, c = (−2 ± √19)/3, about 0.7863 or −2.1196.
6. Keep only values in the interval: −2.1196 is outside (−1, 2), so c = (−2 + √19)/3 ≈ 0.7863.

Why the MVT matters

• If f′(x) = 0 everywhere on an interval, f is constant there.
• If f′(x) > 0 on an interval, f is increasing there; if f′(x) < 0, it's decreasing.
• It bounds how much a function can change: if f(6) = −2 and f′(x) ≤ 10 on [6, 15], then f(15) ≤ −2 + 10 × 9 = 88.

Find c for your own function with the mean value theorem calculator, differentiate with the derivative calculator, or see the secant and tangent lines on the graphing calculator.

Sources: OpenStax Calculus Volume 1, 4.4 The Mean Value Theorem; Paul's Online Notes: The Mean Value Theorem.

Every worked example on this page was checked with Calcist's own engine and an independent reference.