Calculus · step 9 of 13

Mean value & intermediate value theorems

The intermediate value theorem says a continuous function on [a,b] attains every value between f(a) and f(b) — which is how root existence is proved. The mean value theorem says a function continuous on [a,b] and differentiable on (a,b) has some c where f′(c) equals the average rate (f(b)−f(a))/(b−a). Both are existence theorems: they locate no specific point, and their hypotheses must be verified.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Check continuity and differentiabilityIVT needs continuity on the closed interval. MVT needs that plus differentiability on the open interval. State both.
For root existence, find a sign changeIf f(a) < 0 < f(b) and f is continuous, the IVT gives a root in (a,b).
For MVT, compute the average rateThen solve f′(c) = (f(b) − f(a))/(b − a) for c, and check c lies in the open interval.
Note Rolle as the special caseWhen f(a) = f(b), the MVT gives f′(c) = 0 — that is Rolle’s theorem.

Worked example

Find all c satisfying the MVT for f(x) = x² on [1, 3].

f is a polynomial, so it is continuous on [1,3] and differentiable on (1,3) ✓.
Average rate: (f(3) − f(1))/(3 − 1) = (9 − 1)/2 = 4.
f′(x) = 2x, so solve 2c = 4.
c = 2, which lies in (1, 3) ✓.

Answer. c = 2 — the midpoint, as always happens for a quadratic.

Where marks get dropped

These are the specific errors that cost credit on mean value & intermediate value theorems questions — QED's rubric penalises each of them separately.

Applying the MVT to a function with a corner, such as |x| on [−1,1]. Differentiability fails at 0 and the conclusion genuinely fails.
Using the IVT to claim a value is attained outside the range between f(a) and f(b).
Reporting a c outside the open interval. Solutions at the endpoints do not satisfy the theorem’s conclusion.

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Mean value & intermediate value theorems — frequently asked questions

What is Rolle’s theorem?

The MVT with f(a) = f(b): there is a c in (a,b) with f′(c) = 0. Geometrically, a horizontal tangent somewhere between two equal values.

Does the IVT find the root?

No — it proves one exists. Bisection turns the proof into an algorithm by repeatedly halving the interval containing the sign change.

What does the MVT imply about constant functions?

If f′ = 0 everywhere on an interval then f is constant there, since any two points would otherwise give a non-zero average rate. This underpins the +C in integration.

The rest of Calculus

Limits, derivatives, integrals and their applications. Each subtopic below has its own method, worked example and mark-losing traps.

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