The Timpson Lyceum · Calculus 1 (2024)

Extreme Value Theorem and Mean Value Theorem - Lecture 12 (Calculus 1)

Overview

This lecture opens the study of applications of derivatives by asking where a function attains its largest and smallest values. We distinguish absolute (global) extrema from local extrema, see that a function need not have either, and then meet the Extreme Value Theorem, which guarantees both on a closed, bounded interval. We learn that interior extrema occur only at critical points (where the derivative is zero or undefined) — the content of Fermat's Theorem — which gives a calculation-based recipe for finding absolute extrema. The lecture then builds toward the Mean Value Theorem through its special case, Rolle's Theorem, and closes with two important corollaries about functions with zero or equal derivatives. These ideas underpin optimization throughout calculus.

Key Concepts

Absolute (global) extrema

• $f$ has an absolute maximum at a point $c$ in its domain if $f(c) \ge f(x)$ for every $x$ in the domain; an absolute minimum if $f(c) \le f(x)$ for every $x$.
• The extreme value is the output $f(c)$, not the input $c$. A function may have many points achieving the same absolute extremum (e.g. a constant function attains its max and min at every point), or none at all (e.g. $f(x)=x^2$ on $(-\infty,\infty)$ has a min of $0$ but no max).
• Open endpoints can destroy an extremum: $f(x)=x^2$ on $(0,2]$ has no minimum, because $0$ is never attained.

Local extrema

• $f$ has a local maximum at $c$ if $f(x) \le f(c)$ for all $x$ in some open interval containing $c$; a local minimum if $f(x) \ge f(c)$ on such an interval.
• Local extrema describe behavior on a small neighborhood — the "bumps" of the graph. Every absolute extremum (at an interior point) is also a local extremum.

Extreme Value Theorem (EVT)

• If $f$ is continuous on a closed, bounded interval $[a,b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a,b]$.
• Equivalently, there exist $x_1, x_2 \in [a,b]$ with $f(x_1)=m$ the absolute min and $f(x_2)=M$ the absolute max.
• Both hypotheses matter: dropping continuity or using an open/unbounded interval can remove the guarantee. (It is a consequence of compactness; the closed bounded sets of $\mathbb{R}$ are exactly the compact ones.)

Fermat's Theorem (interior extrema have zero derivative)

• If $f$ has a local extremum at an interior point $c$ and $f'(c)$ exists, then $f'(c)=0$.
• Proof sketch: at a local max, the one-sided difference quotients $\dfrac{f(x)-f(c)}{x-c}$ are $\ge 0$ from the left and $\le 0$ from the right; since $f'(c)$ exists, both one-sided limits equal $f'(c)$, forcing $f'(c)=0$.

Critical points

• An interior point $c$ is a critical point if $f'(c)=0$ or $f'(c)$ is undefined (e.g. a corner like $|x|$ at $0$, or a vertical tangent like $\sqrt[3]{x}$ at $0$).
• The converse fails: a critical point need not be an extremum. For $y=x^3$, $f'(0)=0$ but $0$ is neither a max nor a min; for $y=\sqrt[3]{x}$, $f'(0)$ is undefined yet $0$ is no extremum.

Recipe for absolute extrema on $[a,b]$

• Candidates are exactly: the endpoints $a$ and $b$, the interior points where $f'=0$, and the interior points where $f'$ is undefined. Evaluate $f$ at every candidate; the largest output is the absolute max, the smallest is the absolute min.

Rolle's Theorem

• If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $c \in (a,b)$ with $f'(c)=0$.
• Intuition: with equal heights at the ends and no corners, the graph must level off somewhere in between.

Mean Value Theorem (MVT)

• If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ with $f'(c) = \frac{f(b)-f(a)}{b-a}.$
• Intuition (the road trip): if your average velocity over a drive was $60$ mph, then at some instant you were going exactly $60$ mph. The instantaneous slope matches the average slope somewhere.
• Proof idea: subtract the secant line $g(x)$ from $f$ to form $h(x)=f(x)-g(x)$; then $h(a)=h(b)=0$, so Rolle's Theorem gives a $c$ with $h'(c)=0$, i.e. $f'(c)=g'(c)=\dfrac{f(b)-f(a)}{b-a}$.

Corollaries of the MVT

• If $f'(x)=0$ for all $x$ on an interval, then $f$ is constant there (for any two points $x_1,x_2$, the MVT forces $f(x_2)-f(x_1)=f'(c)(x_2-x_1)=0$).
• If $f'(x)=g'(x)$ everywhere on an interval, then $f(x)=g(x)+C$ for some constant $C$ (apply the previous corollary to $h=f-g$).

Worked Example

Find the absolute maximum and minimum of $g(t)=8t - t^4$ on $[-2,1]$.

Use the candidate recipe: evaluate at the endpoints and at the interior critical points (where $g'=0$ or is undefined).

The critical point $t=\sqrt[3]{2}$ lies outside $[-2,1]$, so it is discarded; $g'$ is defined everywhere, so there are no other candidates. Comparing the surviving outputs:

So on $[-2,1]$ the absolute minimum is $-32$ at $t=-2$ and the absolute maximum is $7$ at $t=1$.

Summary

• An absolute extremum is the global high/low value of $f$; a local extremum is high/low only on some small neighborhood. A function may have one, both, or neither.
• The Extreme Value Theorem guarantees an absolute max and min whenever $f$ is continuous on a closed, bounded interval $[a,b]$.
• Fermat's Theorem: at an interior extremum where the derivative exists, $f'=0$; so interior extrema occur only at critical points (where $f'=0$ or is undefined) — but not every critical point is an extremum.
• To find absolute extrema on $[a,b]$, evaluate $f$ at the endpoints and at all interior critical points, then pick the largest and smallest outputs.
• Rolle's Theorem (equal endpoint heights $\Rightarrow$ a flat tangent inside) is the special case used to prove the Mean Value Theorem: $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ for some interior $c$.
• Two MVT corollaries: a function with $f'\equiv 0$ is constant, and two functions with equal derivatives differ only by a constant.