Chapter 1.6, Problem 48E

(a)

To determine

To find: the domain and range of $f$ .

Answer to Problem 48E

The range of graph $f\left(x\right)=x^2$ is $\left(0,\infty \right)$ and range of graph $\left(-\infty ,\infty \right)$ .

Explanation of Solution

Given information:

Given graph of $f$ shown below,

  

Calculation:

The domain is the set of all $x$ values included on graph. Since the curve extends indefinitely in both directions, every value of $x$ includes in the domain.

Therefore the domain of both graphs is $\left(-\infty ,\infty \right)$

The range is the set of all $y$ values included on graph. Therefore the range of graph $f\left(x\right)=x^2$ is $\left(0,\infty \right)$ and range of graph $\left(-\infty ,\infty \right)$ .

(b)

To determine

To find: the $x$ and $y$ intercepts of the graph of $f$ .

Answer to Problem 48E

The $x$ and $y$ intercept of both graph is $\left(0,0\right)$ .

Explanation of Solution

An $x$ intercepts is a point on the graph where $y$ is zero, and an $y$ intercept is a point on the graph where $x$ is zero.

The $x$ and $y$ intercept of both graph is $\left(0,0\right)$ .

(c)

To determine

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

Answer to Problem 48E

  $\left(-\infty ,\infty \right)$

Explanation of Solution

A function is said to be increasing if the $y$ value increases as the $x$ value increases and said to be decreasing as $y$ value decreases the $x$ value increases.

Therefore for graph $f\left(x\right)=x^2$ the increasing interval is $\left[0,\infty \right)$ and decreasing interval is $\left(-\infty ,0\right)$

For $f\left(x\right)=x^2$ the graph is increasing everywhere, so interval is $\left(-\infty ,\infty \right)$ .

(d)

To determine

Whether $f$ is even, odd or neither, then describe the symmetry.

Answer to Problem 48E

Graph of $f\left(x\right)=x^2$ is an even function and graph $f\left(x\right)=x^2$ is an odd function.

Explanation of Solution

Calculation:

A function is even when $f\left(x\right)=f\left(-x\right)$ and odd when $f\left(x\right)=f\left(-x\right)$ for all value of $x$ .

  $\begin{array}{c}f\left(-x\right)={\left(-x\right)}^{2}f\left(-x\right)={\left(-x\right)}^3\\ =x^2=-x^3\\ =f\left(x\right)=-f\left(x\right)\end{array}$

Therefore graph of $f\left(x\right)=x^2$ is an even function and graph $f\left(x\right)=x^2$ is an odd function.