Chapter 2.4, Problem 31AYU

(a)

To determine

To find: the domain of the given function.

Answer to Problem 31AYU

The domain of the given function is the set of all the real numbers.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}-2x+3\text{if}x\ne 0\\ 3x-2\text{if}x=0\end{array}$

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 $f\left(x\right)$ when $x<1$ is given by $-2x+3$ .

In the expression $-2x+3$ is multiplied by $-2$ and 3 is added. These operations can be performed on any real number.

The value of the function for any value of $x\ge 1$ is given by $3x-2$ .

In the expression $3x-2$ , the value of the variable $x$ is multiplied by 3 and then 2 is subtracted from it. These operations can be performed on any real number.

So, the domain of $x$ is the set of all real numbers.

Therefore, the domain of the given function is the set of all the real numbers.

(b)

To determine

To locate: any intercepts of the given function.

Answer to Problem 31AYU

The function does not have any $x$ intercept.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}-2x+3\text{if}x\ne 0\\ 3x-2\text{if}x=0\end{array}$

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.

Determine the points on the graph for which the $x$ coordinate is 0.

The value of the function $f\left(x\right)$ or the $y$ coordinate when $x=0$ is given by $-2x+3$ .

  $\begin{array}{c}f\left(0\right)=-2\left(0\right)+3\\ f\left(0\right)=0+3\\ =3\end{array}$

So, the $y$ intercept is 3.

For finding the $x$ intercept, substitute 0 for $y$ in $f\left(x\right)=-2x+3$ .

  $0=-2x+3$

Add $2x$ to both sides of the equation

  $\begin{array}{c}0+2x=-2x+3+2x\\ 2x=3\end{array}$

Divide both sides of the equation by 2.

  $\begin{array}{c}\frac{2x}{2}=\frac{3}{2}\\ x=\frac{3}{2}\end{array}$

The function $-2x+3$ is defined for $x<1$ . Since $\frac{3}{2}$ is greater than 1, the function does not have any $x$ intercept.

(c)

To determine

To sketch: the graph of the given function.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}-2x+3\text{if}x\ne 0\\ 3x-2\text{if}x=0\end{array}$

Calculation:

Graph each piece to graph the function.

For plotting the graph of the line, $y=-2x+3$ , in the part for which $x<1$ , substitute various values for $x$ in the equation to obtain few points on the graph.

$x$  $y=-2x+3$ $\left(x,y\right)$

0  3  $\left(0,3\right)$ 1   1  $\left(1,1\right)$

In order to plot the graph of the line $3x-2$ in the part for which $x\ge 1$ , substitute various values for $x$ in the equation to obtain few points on the graph.

$x$  $y=3x-2$ $\left(x,y\right)$

1  1  $\left(1,1\right)$ 2  4  $\left(2,4\right)$

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 31AYU

The range of $f$ is $\left\{y|y\ge 1\right\}$ or the interval $\left[1,\infty \right)$ .

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}-2x+3\text{if}x\ne 0\\ 3x-2\text{if}x=0\end{array}$

Calculation:

From the graph, notice that the points on the graph of $f$ have the $y$ coordinates greater than 1, inclusive. For each number $y$ , there is at least one number $x$ in the domain.

So, the range of $f$ is $\left\{y|y\ge 1\right\}$ or the interval $\left[1,\infty \right)$ .

(e)

To determine

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

Answer to Problem 31AYU

The function $f$ is discontinuous on its domain.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}-2x+3\text{if}x\ne 0\\ 3x-2\text{if}x=0\end{array}$

Calculation:

From the graph, it can be observed that there is a discontinuity at $x=0$ . So, the function $f$ is discontinuous on its domain.