AP Calculus AB and BC

Mean Value Theorem vs Intermediate Value Theorem

Q: Which one proves an equation has a solution?

The IVT. Rewrite the equation as f(x) = 0, show f is continuous, and show the endpoint values have opposite signs.

The Intermediate Value Theorem guarantees that a continuous function attains some output value, which is how you prove a root exists. The Mean Value Theorem guarantees that a differentiable function attains some slope, namely the average rate of change across the interval.

Intermediate Value Theorem

Use when: You need to prove an equation has a solution, or that the function equals some specific value somewhere.

Mean Value Theorem

Use when: You need to prove the derivative equals some value, often the average rate of change.

Side by side

	IVT	MVT
Guarantees a value of	$f$	$f'$
Hypotheses	Continuous on $[a,b]$	Continuous on $[a,b]$ and differentiable on $(a,b)$
Conclusion	$f(c) = N$ for some $c$	$f'(c) = \frac{f(b)-f(a)}{b-a}$
Typical use	Proving a root exists	Proving a specific slope occurs

Reading the conclusion is the fastest way to tell them apart. If the statement to prove is about the function's output, reach for the IVT. If it is about a slope, a rate, or a derivative, reach for the MVT.

The MVT asks for more because it promises more. It needs differentiability on the interior, which the IVT never requires, and that extra hypothesis must be stated explicitly for credit.

Both prove existence only

Neither theorem tells you where the point is or how many there are. Naming the theorem and verifying its hypotheses is the entire argument.

Frequently asked questions

Which one proves an equation has a solution?

The IVT. Rewrite the equation as $f(x) = 0$ , show $f$ is continuous, and show the endpoint values have opposite signs.

In the CED: Unit 1: Limits and Continuity, Unit 5: Analytical Applications