AP Calculus AB and BC

Extreme Value Theorem vs MVT

Q: Do I need the function to be differentiable for the Extreme Value Theorem?

Continuity is enough on its own. That is why \lvert x \rvert on [-1,1] has an absolute minimum of 0 at the corner and an absolute maximum of 1 at each endpoint, even though the derivative fails to exist at x = 0.

Q: How is the MVT different from Rolle's theorem?

Rolle's theorem is the MVT with f(a) = f(b). The secant is then horizontal, so the conclusion reads f'(c) = 0. The continuity and differentiability hypotheses are identical and Rolle adds only f(a) = f(b), which is why a function with a corner fails both.

The Extreme Value Theorem needs only continuity on a closed interval and hands you an absolute maximum and an absolute minimum. The Mean Value Theorem also needs differentiability on the open interval, and hands you a point where the instantaneous rate equals the average rate.

Extreme Value Theorem

Use when: You need to know that a largest or smallest value exists before you go hunting for it, and all you can verify is continuity on a closed interval.

Mean Value Theorem

Use when: A question asks you to guarantee some particular value of the derivative, such as an instant when a car was travelling at exactly its average speed.

Side by side

	Extreme Value Theorem	Mean Value Theorem
Hypotheses	$f$ continuous on $[a,b]$	$f$ continuous on $[a,b]$ and differentiable on $(a,b)$
Conclusion	$f$ attains an absolute maximum and an absolute minimum	Some $c$ satisfies $f'(c) = \frac{f(b) - f(a)}{b - a}$
Where the promised point can sit	Anywhere in $[a,b]$ , endpoints included	Interior: the guaranteed $c$ lies in $(a,b)$
How many points promised	Two values, a maximum and a minimum, though one point can serve as both	At least one value of $c$
Applies to $\lvert x \rvert$ on $[-1,1]$	Yes, and the maximum is $1$ , reached at both endpoints	No, the corner at $0$ blocks differentiability

The Extreme Value Theorem is cheap to satisfy and generous with what it returns. Continuity on a closed, bounded interval is the whole hypothesis, and in exchange $f$ genuinely reaches a largest value and a smallest value somewhere on $[a,b]$ . Break either half and the guarantee dissolves. On the open interval $(0,1)$ the function $f(x) = x$ is continuous and reaches neither, because the values it would need sit at the endpoints that are not there.

The Mean Value Theorem costs more. On top of continuity it wants differentiability on the open interval, and it returns a point $c$ inside where the tangent line runs parallel to the secant joining the endpoints. Read as motion, an average speed of $60$ over an hour forces the speedometer to read exactly $60$ at some instant. That extra hypothesis is not decoration: $f(x) = \lvert x \rvert$ on $[-1,1]$ is continuous, so the EVT applies, but the secant slope is $0$ and $f'$ only ever takes the values $-1$ and $1$ .

\text{EVT: } f(c_1) \le f(x) \le f(c_2) \ \text{ for all } x \in [a,b] \qquad \text{MVT: } f'(c) = \frac{f(b) - f(a)}{b - a}

Cite the hypotheses that belong to the theorem you named

The characteristic error is a justification that claims one theorem's conclusion under the other one's name: asserting that a maximum exists by the Mean Value Theorem, or that some value of $f'$ is achieved by the Extreme Value Theorem. Continuity alone buys extrema and says nothing about the derivative. Differentiability buys the matching rate and says nothing about which value is largest. Both are existence statements, so neither one locates its point, and a question that asks where the extremum is still needs the candidates test.

Frequently asked questions

Does the Extreme Value Theorem tell me where the maximum is?

The theorem locates nothing. It promises only that a maximum is attained somewhere on $[a,b]$ , and finding it is a separate job: list the critical numbers and the two endpoints, evaluate $f$ at each, and compare.

Do I need the function to be differentiable for the Extreme Value Theorem?

Continuity is enough on its own. That is why $\lvert x \rvert$ on $[-1,1]$ has an absolute minimum of $0$ at the corner and an absolute maximum of $1$ at each endpoint, even though the derivative fails to exist at $x = 0$ .

How is the MVT different from Rolle's theorem?

Rolle's theorem is the MVT with $f(a) = f(b)$ . The secant is then horizontal, so the conclusion reads $f'(c) = 0$ . The continuity and differentiability hypotheses are identical and Rolle adds only $f(a) = f(b)$ , which is why a function with a corner fails both.

In the CED: Unit 5: Analytical Applications