Chapter 4.3, Problem 50E

a.

To determine

To find the absolute maximum and absolute minimum.

Answer to Problem 50E

The absolute maximum is $(0,2)$ and the absolute minimum at $(2,-1)$ and $(-2,-1)$ .

Explanation of Solution

Given:

$x$	0	1	2
$f$	2	0	-1
$f\text{'}$	Does not exist	0	Does not exist
$f\text{'}\text{'}$	Does not exist	0	Does not exist

$x$	0<x<1	1<x<2	2<x<3
$f$	+	-	-
$f\text{'}$	-	-	+
$f\text{'}\text{'}$	+	-	-

Calculation:

The absolute maximum is where the $f\text{'}\text{'}$ decreasing and $f\text{'}\text{'}$ increasing.

From the given information,

The absolute maximum is $(0,2)$ and the absolute minimum at $(2,-1)$ and $(-2,-1)$ .

b.

To determine

To find the point of inflection.

Answer to Problem 50E

The point of inflection is $(1,0)$ and $(-1,0)$ .

Explanation of Solution

Given:

$x$	0	1	2
$f$	2	0	-1
$f\text{'}$	Does not exist	0	Does not exist
$f\text{'}\text{'}$	Does not exist	0	Does not exist

$x$	0<x<1	1<x<2	2<x<3
$f$	+	-	-
$f\text{'}$	-	-	+
$f\text{'}\text{'}$	+	-	-

Calculation:

The point of inflection is the point where $f\text{'}=0$ .

From the given information,

The point of inflection is $(1,0)$ and $(-1,0)$ .

c.

To determine

The find the graph.

Answer to Problem 50E

The graph is as follow:

  

Explanation of Solution

Given:

$x$	0	1	2
$f$	2	0	-1
$f\text{'}$	Does not exist	0	Does not exist
$f\text{'}\text{'}$	Does not exist	0	Does not exist

$x$	0<x<1	1<x<2	2<x<3
$f$	+	-	-
$f\text{'}$	-	-	+
$f\text{'}\text{'}$	+	-	-

Calculation:

From the graph,

The end points are $(-3,0)$ , $(3,0)$ .

There is no straight line in the graph.

The resulting graph will be

  

d.

To determine

To find the summary of the $f(3)$ and $f(-3)$ .

Answer to Problem 50E

  $f(3)=f(-3)$

Explanation of Solution

Given:

$x$	0	1	2
$f$	2	0	-1
$f\text{'}$	Does not exist	0	Does not exist
$f\text{'}\text{'}$	Does not exist	0	Does not exist

$x$	0<x<1	1<x<2	2<x<3
$f$	+	-	-
$f\text{'}$	-	-	+
$f\text{'}\text{'}$	+	-	-

Calculation:

  $f$ is even, $f(3)=f(-3)$ .

By the continuity of $f$ , since $f(x)<0$ at $2<x<3$ .

At $f(3)\le 0$ , since $f(2)=-1$ and $f\text{'}(x)>0$ when $2<x<3$

At $f(3)>-1$ .

At $f(3)=f(-3)$ , then $-1<f(x)\le 0$ and $-1<f(-3)\le 0$ .