5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

Topic 5.1's Mean Value Theorem guarantees that a certain rate occurs somewhere, without saying where. The Extreme Value Theorem is the same kind of existence theorem for values: under the right conditions, a function must reach a highest and a lowest value. This page states those conditions, separates global from local extrema, and introduces critical points, the only interior locations where an extremum can occur.

5.2 The Extreme Value Theorem: proving extrema exist

The Extreme Value Theorem (EVT) states:

If a function $f$ is continuous over the closed interval $[a,b]$, then the Extreme Value Theorem guarantees that $f$ has at least one minimum value and at least one maximum value on $[a,b]$.

This is the exact hypothesis in FUN-1.C.1 and the core of learning objective FUN-1.C: Justify conclusions about functions by applying the Extreme Value Theorem. The phrase closed interval is essential: both endpoints must be included.

A function may fail to attain an extreme value if continuity or closedness is missing. For example, $f(x)=\dfrac{1}{x}$ is not continuous on an interval containing $x=0$, and $f(x)=x$ on the open interval $(0,1)$ has neither an attained absolute minimum nor an attained absolute maximum.

Global, local, and critical points

A global (absolute) maximum on a domain is a function value at least as large as every other value in that domain. A global (absolute) minimum is at most every other value. A local (relative) maximum is highest only within a nearby interval, while a local (relative) minimum is lowest nearby.

A critical point occurs at a point in the domain where

$$ f'(x)=0 $$

or where $f'(x)$ does not exist. This definition is FUN-1.C.2. The corresponding $x$-value is often called a critical number.

FUN-1.C.3 states that all local (relative) extrema occur at critical points, although not all critical points are local extrema. For instance, $f(x)=x^3$ has $f'(0)=0$, but the function continues increasing through $x=0$; therefore, $x=0$ is critical but is not a local maximum or minimum.

Sign-change insight: If $f'$ changes from negative to positive at $x=2$, the function changes from decreasing to increasing, so $f$ has a local minimum at $x=2$.

Absolute extrema on a closed interval

To locate an absolute minimum or maximum on $[a,b]$, evaluate $f$ at every relevant interior critical point and at both endpoints. The endpoint check is indispensable: a global extreme can occur where the derivative is not zero because endpoints are not interior points.

Worked example. Find the absolute minimum of

$$ f(x)=x^2-4x+5 $$

on $[0,5]$. First,

$$ f'(x)=2x-4, $$

so the critical point is $x=2$. Compare:

$$ f(0)=5,\qquad f(2)=1,\qquad f(5)=10. $$

Thus the absolute minimum value is

$$ \boxed{1} $$

at $x=2$. The conclusion is justified because the polynomial is continuous on $[0,5]$, so the EVT guarantees that an absolute minimum exists, and the critical-point/endpoints comparison identifies it.

Retrieval check. A function is continuous on $[-3,4]$, and $f'(x)$ is undefined at $x=1$ but defined elsewhere. What points must be checked when searching for an absolute minimum? Answer: $x=-3$, $x=1$, and $x=4$, provided $x=1$ lies in the function’s domain.