Chapter 2.4, Problem 37AYU

(a)

To determine

To find: the domain of the given function.

Answer to Problem 37AYU

The domain of the given function is $\left\{x|x\ge -2,x\ne 0\right\}$ or the interval $\left[-2,0\right)\cup \left(0,\infty \right)$ .

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}\left|x\right|\text{\hspace{1em}if}-2\le x<0\\ x^3\text{\hspace{1em}if}x>0\end{array}$

Calculation:

The domain of the function $f\left(x\right)$ , is the set of all the possible values of $x$ .

The value of the function $f\left(x\right)$ or the $y$ coordinate when $x$ lies between 0 and $-2$ , inclusive, is given by $\left|x\right|$ .

The value of the function $f\left(x\right)$ when $x>0$ is given by $x^3$ .

Since the value of the function $x=0$ is not defined, the domain of $x$ is the set of all the real numbers greater than or equal to $-2$ excluding 0.

Therefore, the domain of the given function is $\left\{x|x\ge -2,x\ne 0\right\}$ or the interval $\left[-2,0\right)\cup \left(0,\infty \right)$

(b)

To determine

To locate: any intercepts of the given function.

Answer to Problem 37AYU

The given function has no intercepts.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}\left|x\right|\text{\hspace{1em}if}-2\le x<0\\ x^3\text{\hspace{1em}if}x>0\end{array}$

Calculation:

The $x$ intercepts are those points for which $y$ coordinate is zero and the $y$ intercepts are those points for which the $x$ coordinate is zero.

Since the graph does not define any value for the functionwhen $x=0$ , there are no $y$ intercepts for the graph.

There are no $x$ intercepts for the graph since, the value of the function does not equal zero in the defined functions.

Therefore, the given function has no intercepts.

(c)

To determine

To sketch: the graph of the given function.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}\left|x\right|\text{\hspace{1em}if}-2\le x<0\\ x^3\text{\hspace{1em}if}x>0\end{array}$

Calculation:

Graph each piece to graph the function.

For plotting the graph of the absolute function, $f\left(x\right)=\left|x\right|$ , in the part for which $-2\le x<0$ , substitute various values for $x$ in the equation to obtain few points on the graph.

$x$   $y=\left|x\right|$$\left(x,y\right)$

1   1 $\left(1,1\right)$   $-2$ 2 $\left(-2,2\right)$

In order to plot the graph of the line $f\left(x\right)=x^3$ in the part for which $x>0$ , substitute various values for $x$ in the equation to obtain few points on the graph.

$x$   $y=x^3$ $\left(x,y\right)$

1   1 $\left(1,1\right)$

Plot the points and draw the lines to get the graph of the function.

  

(d)

To determine

To find: the range based on the graph.

Answer to Problem 37AYU

The range of $f$ is $\left\{y|y>0\right\}$ or the interval $\left(0,\infty \right)$ .

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}\left|x\right|\text{\hspace{1em}if}-2\le x<0\\ x^3\text{\hspace{1em}if}x>0\end{array}$

Calculation:

From the graph, notice that the points on the graph of $f$ have the $y$ coordinatesgtreater than 0. For each number $y$ , there is at least one number $x$ in the domain.

So, the range of $f$ is $\left\{y|y>0\right\}$ or the interval $\left(0,\infty \right)$ .

(e)

To determine

To find: whether $f$ continuous on its domain or not.

Answer to Problem 37AYU

The function $f$ is continuous on its domain.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}\left|x\right|\text{\hspace{1em}if}-2\le x<0\\ x^3\text{\hspace{1em}if}x>0\end{array}$

Calculation:

From the graph, it can be observed the graph is continuous at $x=0$ . So, the function $f$ is continuous on its domain.

  $\begin{array}{c}f\left(0^{-}\right)=\left|0\right|\\ =0\end{array}$

And

  $\begin{array}{c}f\left(0\right)=0^3\\ =0\end{array}$

And

  $\begin{array}{c}f\left(0^{+}\right)=0^3\\ =0\end{array}$

Hence

  $f\left(0^{+}\right)=f\left(0\right)=f\left(0^{-}\right)$