Chapter 2.4, Problem 33AYU

(a)

To determine

To find: the domain of the given function.

Answer to Problem 33AYU

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

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}x+3\text{\hspace{1em}if}-2\le x<1\\ 5\text{\hspace{1em}if}x=1\\ -x+2\text{\hspace{1em}if}x>1\end{array}$

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 $-2\le x<1$ is given by $x+3$ .

In the expression $x+3$ , $x$ is added to 3. These operations can be performed on any real number.

The value of the function or the $y$ coordinate when $x=1$ is 5.

The value of the function $f\left(x\right)$ when $x>1$ is given by $-x+2$ .

In the expression $-x+2$ , the value of the opposite of the variable $x$ is added by 2. 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 33AYU

The intercepts are $\left(0,3\right)$ and $\left(2,0\right)$ .

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}x+3\text{\hspace{1em}if}-2\le x<1\\ 5\text{\hspace{1em}if}x=1\\ -x+2\text{\hspace{1em}if}x>1\end{array}$

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 $x+3$ .

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

So, the $y$ intercept is 3.

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

Consider the function $f\left(x\right)=-x+2$ . If the $y$ coordinate is 0, then $f\left(x\right)=0$ .

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

So, the $x$ intercept is $\left(2,0\right)$ .

Therefore, the intercepts are $\left(0,3\right)$ and $\left(2,0\right)$ .

(c)

To determine

To sketch: the graph of the given function.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}x+3\text{\hspace{1em}if}-2\le x<1\\ 5\text{\hspace{1em}if}x=1\\ -x+2\text{\hspace{1em}if}x>1\end{array}$

Calculation:

Graph each piece to graph the function.

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

$x$  $y=x+3$ $\left(x,y\right)$

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

In order to plot the graph of the line $y=-x+2$ in the part for which $x>1$ , substitute various values for $x$ in the equation to obtain few points on the graph.

$x$  $y=-x+2$ $\left(x,y\right)$

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 33AYU

The range of $f$ is $\left\{y|y<4,y=5\right\}$ or the interval $\left(-\infty ,4\right)\cup \left\{5\right\}$ .

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}x+3\text{\hspace{1em}if}-2\le x<1\\ 5\text{\hspace{1em}if}x=1\\ -x+2\text{\hspace{1em}if}x>1\end{array}$

Calculation:

From the graph, notice that the points on the graph of $f$ have the $y$ coordinates less than 4, inclusive. For each number $y$ , there is at least one number $x$ in the domain $\left(1,5\right)$ is a point on the graph.

So, the range of $f$ is $\left\{y|y<4,y=5\right\}$ or the interval $\left(-\infty ,4\right)\cup \left\{5\right\}$ .

(e)

To determine

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

Answer to Problem 33AYU

The function $f$ is discontinuous on its domain.

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\{\begin{array}{l}x+3\text{\hspace{1em}if}-2\le x<1\\ 5\text{\hspace{1em}if}x=1\\ -x+2\text{\hspace{1em}if}x>1\end{array}$

Calculation:

From the graph, it can be observed that there is a discontinuity at $x=1$ . So, the function $f$ is discontinuous on its domain.