Chapter 1.5, Problem 92E

a.

To determine

Find the domain and range of $f$ .

Answer to Problem 92E

Domain $\left[-4,5\right]$ ,Range $\left[0,9\right]$

Explanation of Solution

Calculation:

Let us consider the following function

  

From the graph we can see that there is a closed and an open dot.

The open dot that is, $\left(5,9\right)$ is in the first quadrant.

This implies $x=5$ which is not included in the domain.

The closed dot this is, $\left(-4,-1\right)$ is in the second quadrant.

This implies $x=-4$ which is included in the domain.

The graph does not extend beyond these points.

Hence, the domain becomes $\left[-4,5\right]$

We now find the range of the function.

As the point $\left(0,9\right)$ is on the graph

This implies $f(0)=9$ .

Also, the point $(3,0)$ is on the graph.

This implies $f(3)=0$ .

The graph does not extend beyond these points.

Hence, the range beccomes $\left[0,9\right]$

b.

To determine

Find the zero(s) of $f$ .

Answer to Problem 92E

  $3$

Explanation of Solution

Calculation:

Let us consider the following function

  

We know zeros are those values of $x$ for which $f(x)=0$

From the graph, we see that when $x=3$ , then $y=0$

That is, $f(3)=0$

Hence the zeros of the function are $3$ .

c.

To determine

Determine the open intervals on which $f$ is increasing, decreasing, or constant.

Answer to Problem 92E

The function is increasing over the interval $\left(-4,0\right)$ and $\left(3,5\right)$ .

The function decreases over the interval $\left(0,3\right)$ .

Explanation of Solution

Calculation:

Let us consider the following function

  

Let $x$ and $y$ belong to any interval.

Without loss of generality, we assume $x<y$

If a function $f$ is constant, then $f(x)=f(y)$

If a function $f$ is increasing, then $f(x)<f(y)$

If a function $f$ is decreasing, then $f(x)>f(y)$

Using above definitions, we conclude that

Hence, the function is increasing over the interval $\left(-4,0\right)$ and $\left(3,5\right)$

Hence, the function decreases over the interval $\left(0,3\right)$ .

d.

To determine

Approximate relative minimum value of $f$ .

Answer to Problem 92E

  $\left(0,3\right)$

Explanation of Solution

Calculation:

Let us consider the following function

  

We consider an interval $\left(x,y\right)$ .

Let $p$ belongs to interval $\left(x,y\right)$ such that $x<p<y$

If a function $f(p)$ is a relative minimum, then $f(p)\le f(x)$

If a function $f(p)$ is a relative maximum then, $f(p)\ge f(x)$

Using above definitions and the graph, we conclude that

Hence, the relative minimum point is approximate at $\left(0,3\right)$ .

e.

To determine

Is $f$ even, odd, or neither?

Answer to Problem 92E

It is neither odd nor even function.

Explanation of Solution

Calculation:

Let us consider the following function

  

A function $f$ is even only if $f\left(-x\right)=f\left(x\right)$ . An even function is symmetric with respect to $y-axis$ .

A function $f$ is odd only if $f\left(-x\right)=-f\left(x\right)$ . An odd function is symmetric with respect to origin.

As the function is not symmetric,

Hence, it is neither odd nor even function.