Introduction

This topic focuses on finding absolute (global) extrema on closed intervals using the Candidates Test. The essential knowledge tells us that absolute (global) extrema of a function on a closed interval can only occur at critical points or at endpoints. This systematic approach guarantees we find the highest and lowest values of a function on a given interval.

The Candidates Test

Why Absolute Extrema Occur Where They Do

On a closed interval $[a,b]$, a continuous function must have absolute maximum and minimum values (by the Extreme Value Theorem). These extrema can only occur at:

The Candidates Test Procedure

Step 1: Find all critical points of $f$ in the open interval $(a,b)$

Step 2: Evaluate $f$ at each critical point and at both endpoints

Step 3: Compare all values:

Example 1:

Find the absolute extrema of $f(x)=x^3-3x^2+1$ on $[-1,3]$.

Find critical points:

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

$x=0$ and $x=2$ (both in $(-1,3)$)

Evaluate at candidates:

$f(-1)=-1-3+1=-3$

$f(0)=1$

$f(2)=8-12+1=-3$

$f(3)=27-27+1=1$

Absolute maximum: $1$ at $x=0$ and $x=3$

Absolute minimum: $-3$ at $x=-1$ and $x=2$

When Critical Points Are Outside the Interval

Checking Domain Restrictions

If a critical point is not in the interval $[a,b]$, it cannot be an absolute extremum on that interval.

Example 2:

Find the absolute extrema of $g(x)=x^2-4x+5$ on $[0,2]$.

Find critical points:

$g^{\prime }(x)=2x-4$

$2x-4=0$

$x=2$ (this is an endpoint, not in the interior)

Actually, $x=2$ is the right endpoint. No critical points in $(0,2)$.

Evaluate at candidates:

$g(0)=5$

$g(2)=4-8+5=1$

Absolute maximum: $5$ at $x=0$

Absolute minimum: $1$ at $x=2$

Note: Both extrema occur at endpoints.

Functions with Multiple Critical Points

Example 3:

Find the absolute extrema of $h(x)=x^4-4x^3+2$ on $[-1,4]$.

Find critical points:

$h^{\prime }(x)=4x^3-12x^2=4x^2(x-3)$

$x=0$ and $x=3$ (both in $(-1,4)$)

Evaluate at candidates:

$h(-1)=1+4+2=7$

$h(0)=2$

$h(3)=81-108+2=-25$

$h(4)=256-256+2=2$

Absolute maximum: $7$ at $x=-1$

Absolute minimum: $-25$ at $x=3$

Trigonometric Functions on Closed Intervals

Example 4:

Find the absolute extrema of $f(x)=2\sin (x)+\cos (2x)$ on $[0,\pi ]$.

Find critical points:

$f^{\prime }(x)=2\cos (x)-2\sin (2x)=2\cos (x)-4\sin (x)\cos (x)$

$f^{\prime }(x)=2\cos (x)(1-2\sin (x))$

Set equal to zero:

$\cos (x)=0$ gives $x=\frac{\pi }{2}$

$1-2\sin (x)=0$ gives $\sin (x)=\frac{1}{2}$, so $x=\frac{\pi }{6}$ or $x=\frac{5\pi }{6}$

Critical points in $[0,\pi ]$: $x=\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}$

Evaluate at candidates:

$f(0)=0+1=1$

$f\left(\frac{\pi }{6}\right)=2\left(\frac{1}{2}\right)+\cos \left(\frac{\pi }{3}\right)=1+\frac{1}{2}=\frac{3}{2}$

$f\left(\frac{\pi }{2}\right)=2(1)+\cos (\pi )=2-1=1$

$f\left(\frac{5\pi }{6}\right)=2\left(\frac{1}{2}\right)+\cos \left(\frac{5\pi }{3}\right)=1+\frac{1}{2}=\frac{3}{2}$

$f(\pi )=0+1=1$

Absolute maximum: $\frac{3}{2}$ at $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$

Absolute minimum: $1$ at $x=0,\frac{\pi }{2},\pi$

Rational Functions with Restrictions

Example 5:

Find the absolute extrema of $g(x)=\frac{x}{x^2+1}$ on $[-2,3]$.

Find critical points:

$g^{\prime }(x)=\frac{(x^2+1)(1)-x(2x)}{(x^2+1{)}^2}=$

$\frac{x^2+1-2x^2}{(x^2+1{)}^2}=$

$\frac{1-x^2}{(x^2+1{)}^2}$

Set numerator equal to zero: $1-x^2=0$ $x=\pm 1$

Both $x=-1$ and $x=1$ are in $[-2,3]$.

Evaluate at candidates:

$g(-2)=\frac{-2}{5}=-0.4$

$g(-1)=\frac{-1}{2}=-0.5$

$g(1)=\frac{1}{2}=0.5$

$g(3)=\frac{3}{10}=0.3$

Absolute maximum: $\frac{1}{2}$ at $x=1$

Absolute minimum: $-\frac{1}{2}$ at $x=-1$

Practice Section