Introduction

This topic explores extrema (maximum and minimum values) of functions and introduces critical points. The essential knowledge tells us that if a function $f$ is continuous over the interval $[a,b]$, then the Extreme Value Theorem guarantees that $f$ has at least one minimum value and at least one maximum value on $[a,b]$, a point on a function where the first derivative equals zero or fails to exist is a critical point of the function, and all local (relative) extrema occur at critical points of a function, though not all critical points are local extrema. Understanding these concepts is fundamental to analyzing function behavior.

The Extreme Value Theorem

Statement of the Theorem

Extreme Value Theorem (EVT): If $f$ is continuous on the closed interval $[a,b]$, then $f$ has both:

on that interval.

Key Point: The EVT guarantees existence but does not tell us where these extrema occur or what their values are.

Real-World Example: Imagine the graph is like a hill. no matter if you're going up or down there's always a time where you're higher than any other time, same for the bottom.

Why Continuity and Closed Intervals Matter

The EVT requires BOTH conditions:

Example 1:

Consider $f(x)=x^2$ on $[0,3]$.

$f$ is continuous on $[0,3]$ (polynomial) ✓

By the EVT, $f$ has a maximum and minimum on $[0,3]$.

Minimum: $f(0)=0$ at $x=0$

Maximum: $f(3)=9$ at $x=3$

When the EVT Does Not Apply

Example 2:

For $f(x)=\frac{1}{x}$ on $[0,2]$:

$f$ is not continuous at $x=0$ (undefined there).

The EVT does not apply. Indeed, $f$ has no maximum value on this interval (it approaches $\infty$ as $x\to 0^{+}$).

Example 3:

For $f(x)=x$ on $(0,1)$ (open interval):

$f$ is continuous, but the interval is not closed.

The EVT does not apply. The function has no maximum or minimum on the open interval (it approaches but never reaches $0$ and $1$).

Critical Points

Definition of Critical Points

A critical point of $f$ is a value $x=c$ in the domain where:

Important: Critical points are $x$-values, not points $(x,y)$.

Finding Critical Points

Example 4:

Find all critical points of $f(x)=x^3-3x^2-9x+5$

Find $f^{\prime }(x)$: $f^{\prime }(x)=3x^2-6x-9$

Set $f^{\prime }(x)=0$: $3x^2-6x-9=0$

$x^2-2x-3=0$

$(x-3)(x+1)=0$

$x=3$ or $x=-1$

The derivative exists everywhere, so the critical points are $x=-1$ and $x=3$.

You can see patterns using the graph. Look where the visible changes in the graph are, and then look at the critical points.

I highly recommend you to graph the following examples to see the visuals for yourself.

Example 5:

Find all critical points of $f(x)=x^{2/3}$.

Find $f^{\prime }(x)$: $f^{\prime }(x)=\frac{2}{3}{x}^{-1/3}=\frac{2}{3x^{1/3}}$

Set $f^{\prime }(x)=0$: $\frac{2}{3x^{1/3}}=0$ has no solution

Check where $f^{\prime }(x)$ is undefined: $f^{\prime }(x)$ is undefined at $x=0$

The critical point is $x=0$.

Global (Absolute) Extrema

Definition

Global Maximum: $f(c)$ is a global maximum if $f(c)\ge f(x)$ for all $x$ in the domain.

Global Minimum: $f(c)$ is a global minimum if $f(c)\le f(x)$ for all $x$ in the domain.

Finding Global Extrema on Closed Intervals

For a continuous function on $[a,b]$, global extrema occur at:

The Candidates Test:

Example 6: 

This graph shows how to find global (absolute) extrema on a closed interval.

The function has critical points at $x=1$ and $x=3$ (mark these with points in Desmos). To find global extrema, we must check: (1) the critical points $x=1$ and $x=3$, and (2) the endpoints $x=0$ and $x=5$.

Evaluating:

$g(0)=2$

$g(1)=6$

$g(3)=2$

$g(5)=12$

The global maximum is $12$ at $x=5$ (an endpoint!), and the global minimum is $2$, occurring at both $x=0$ and $x=3$. This demonstrates that global extrema can occur at either critical points or endpoints.

Local (Relative) Extrema

Definition

Local Maximum: $f(c)$ is a local maximum if $f(c)\ge f(x)$ for all $x$ near $c$.

Local Minimum: $f(c)$ is a local minimum if $f(c)\le f(x)$ for all $x$ near $c$.

Local extrema are "peaks and valleys" - they don't have to be the highest or lowest points overall.

Key Theorem About Local Extrema

If $f$ has a local extremum at $x=c$, then $c$ is a critical point.

This means: All local extrema occur at critical points.

However: Not all critical points are local extrema!

Example 7:

For $f(x)=x^3$:

Find critical points: $f^{\prime }(x)=3x^2$

$3x^2=0$

$x=0$

At $x=0$, the function has a critical point, but it's neither a local maximum nor minimum (it's an inflection point).

Distinguishing Local from Global

A global extremum is also a local extremum, but a local extremum may not be global.

Example 8:

For $f(x)=x^3-3x$ on $[-2,2]$:

Critical points: $f^{\prime }(x)=3x^2-3=0$

$x=\pm 1$

Evaluate:

$f(-2)=-8+6=-2$

$f(-1)=-1+3=2$ (local maximum)

$f(1)=1-3=-2$ (local minimum)

$f(2)=8-6=2$

At $x=-1$: local maximum of $2$ (tied for global maximum)

At $x=1$: local minimum of $-2$ (tied for global minimum)

Analyzing Critical Points

Types of Critical Points

A critical point can be:

Editor's Note: A saddle point in this context occurs when there is an inflection point with a horizontal tangent, eg $x=0$ on $f(x)=x^3$.

Example 9: 

Find critical points: $f^{\prime }(x)=4x^3-12x^2=4x^2(x-3)$

$x=0$ or $x=3$

Analyze each:

Using Test Points

To determine if a critical point is a local extremum, check the sign of $f^{\prime }(x)$ around it:

Local Maximum: $f^{\prime }$ changes from positive to negative

Local Minimum: $f^{\prime }$ changes from negative to positive

Neither: $f^{\prime }$ doesn't change sign

Example 10:

For $f(x)=x^3-3x^2-9x+5$ with critical points $x=-1$ and $x=3$:

$f^{\prime }(x)=3x^2-6x-9$

Test intervals:

At $x=-1$: $f^{\prime }$ changes from $+$ to $-$, so local maximum

At $x=3$: $f^{\prime }$ changes from - to +, so local minimum

Justifying Extrema Claims

Complete Justification Process

To justify that a value is an extremum:

Example 11:

Justify that $f(x)=2x^3-15x^2+24x$ has a local maximum at $x=1$ on the interval $[0,5]$.

Find critical points: $f^{\prime }(x)=6x^2-30x+24=6(x^2-5x+4)=6(x-1)(x-4)$

$x=1$ or $x=4$

Test signs around $x=1$:

Since $f^{\prime }$ changes from positive to negative at $x=1$, the function has a local maximum at $x=1$.

Practice Section