Theorems, hypotheses, and critical points

Understand the idea

The Mean Value Theorem requires continuity on a closed interval and differentiability on its interior; it guarantees an interior tangent matching the secant slope. The Extreme Value Theorem needs continuity on a closed bounded interval and guarantees at least one absolute minimum and one absolute maximum. Critical numbers are interior domain points where the derivative is zero or does not exist; they are candidates, not automatically extrema.

f'(c)=\frac{f(b)-f(a)}{b-a}\quad\text{for some }c\in(a,b)\text{ under MVT hypotheses}

Worked example

Apply the MVT to $\displaystyle f(x)=x^2$ on $\displaystyle [1,3]$ .

A polynomial satisfies both hypotheses. The secant slope is

\displaystyle (9-1)/(3-1)=4

. Solving

\displaystyle f'(c)=2c=4

gives

\displaystyle c=2

, which lies inside

\displaystyle (1,3)

.

Try it yourself

Does continuity on $\displaystyle (0,1)$ alone guarantee an absolute maximum?

Show solution and reasoning

No. For example $\displaystyle f(x)=x$ is continuous there but never reaches $\displaystyle 1$ . Closed endpoints are needed for the EVT guarantee.