The Mean Value Theorem

You’ll be able to

State the hypotheses and conclusion of the Mean Value Theorem
Find the guaranteed value c for a given function and interval
Interpret the MVT as an average rate equaling an instantaneous rate

Average slope must occur somewhere

The Mean Value Theorem (MVT) says: if f is continuous on [a, b] and differentiable on (a, b), then there is at least one point c in (a, b) where the instantaneous rate f′(c) equals the average rate over the interval. Geometrically, somewhere the tangent line is parallel to the secant connecting the endpoints. Physically, if your average speed on a trip was 60 mph, at some instant your speedometer read exactly 60.

Mean Value Theorem

f′(c) = [ f(b) − f(a) ] / (b − a) for some c in (a, b)

The right side is the secant slope (average rate); f′(c) is the tangent slope (instantaneous rate). MVT guarantees they match at some c.

Both hypotheses matter

The MVT needs continuity on the closed interval and differentiability on the open interval. Continuity keeps the graph connected; differentiability rules out corners and vertical tangents where no single tangent slope exists. If either fails — say f has a sharp corner — the conclusion can break down. Rolle’s Theorem is the special case where f(a) = f(b), forcing f′(c) = 0 (a horizontal tangent) somewhere between.

Worked example

For f(x) = x² on [1, 3], find the value c guaranteed by the Mean Value Theorem.

1.f is a polynomial, so it is continuous on [1, 3] and differentiable on (1, 3) — hypotheses met.
2.Compute the average rate: [f(3) − f(1)]/(3 − 1) = (9 − 1)/2 = 8/2 = 4.
3.Set the instantaneous rate equal to it: f′(c) = 2c = 4.
4.Solve: c = 2, which lies in (1, 3) as required.

Answer: c = 2. At x = 2 the tangent slope f′(2) = 4 equals the average slope of the function across [1, 3].

Watch out

The MVT requires differentiability on the open interval and continuity on the closed interval — the two hypotheses are not the same. A function with a corner (like |x|) is continuous but not differentiable there, so the MVT need not apply.

Checkpoint

A car travels 120 miles in 2 hours. What does the Mean Value Theorem guarantee, assuming its position is differentiable?

Checkpoint

For f(x) = x² − 4x on [0, 4], find the c value from the Mean Value Theorem.

On the exam

MVT justifications must state both hypotheses ("f is continuous on [a,b] and differentiable on (a,b)") before invoking the conclusion. Many free-response points hinge on that explicit verification, not on solving for c.

Answer the 2 checkpoints as you read.