Chapter 2.4, Problem 38AYU

(a)

To determine

To find: the domain of the given function.

Answer to Problem 38AYU

The domain of the given function is the set of all the real numbers greater than equal to $0$

Explanation of Solution

Given information:

Given function

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

Calculation:

From the given interval, the domain of $f\left(x\right)=2-x$ is the set of all real numbers greater than or equal to $-3$ and less than 1. The domain of $f\left(x\right)=\sqrt{x}$ is the set of all real; numbers greater than 0.

Thus, the domain of $f$ is the set of all real numbers greater than or equal to $0$ .

(b)

To determine

To locate: any intercepts of the given function.

Answer to Problem 38AYU

The $y$ intercept of the given function is 2.

  $x$ intercepts are 2 and 0.

Explanation of Solution

Given information:

Given function

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

Calculation:

The $y$ intercept is the value of the function when $x$ is 0.

When $x=0$ , the equation for $f$ is given by $f\left(x\right)=2-x$ .

Substitute 0 for $x$ in the function.

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

The $y$ intercept of the given function is 2.

Find the $x$ intercept.

Substitute 0 for $f\left(x\right)$ in the given functions.

  $2-x=0$  and  $\sqrt{x}=0$ .

Solve for $x$ .

  $x=2$ and $x=0$

The $x$ intercepts are 2 and 0.

(c)

To determine

To sketch: the graph of the given function.

Explanation of Solution

Given information:

Given function

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

Calculation:

First, graph the function $f\left(x\right)=2-x$ . Since $-3\le x<1$ , we need to find points for which $x$ is greater than or equal to $-3$ and less than 1.

Choose any value of $x$ say, $-2$ and find the corresponding $y$ value.

  $\begin{array}{c}f\left(-2\right)=2-\left(-2\right)\\ =4\end{array}$

Thus, one point is $\left(-2,4\right)$ .

Choose some for value for $x$ and create a table.

$x$	$y=f\left(x\right)=2-x$	$\left(x,y\right)$
$-3$	5	$\left(-3,5\right)$
$-2$	4	$\left(-2,4\right)$
$-1$	3	$\left(-1,3\right)$
0	2	$\left(0,2\right)$

Plot the points on a coordinate plane and join them using straight line, Since the point $\left(1,1\right)$ is not included in the graph, use an open circle to denote that point.

  

Similarly, graph the function $f\left(x\right)=\sqrt{x}$ and find points for which $x$ is greater than 1.

Create a table.

$x$	$y=f\left(x\right)=\sqrt{x}$	$\left(x,y\right)$
$2$	1.41	$\left(2,1.41\right)$
$4$	2	$\left(4,2\right)$
$5$	2.23	$\left(5,2.23\right)$

Plot the points on the same coordinate plane and join them using a straight line.

  

(d)

To determine

To find: the range based on the graph.

Answer to Problem 38AYU

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

Explanation of Solution

Given information:

Given function

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

Calculation:

From the graph, the range of the given functions as the set of real numbers greater than 1. Thus, the range of $f$ is $\left\{y|y>1\right\}$ or the interval $\left(1,\infty \right)$ .

(e)

To determine

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

Answer to Problem 38AYU

The function $f$ is discontinuous on its domain.

Explanation of Solution

Given information:

Given function

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

Calculation:

Now, at $x<=1$ .

  $\begin{array}{c}f\left(1^{-}\right)=2-x\\ =2-1\\ =1\end{array}$

Now, at $x>1$ .

  $\begin{array}{c}f\left(1^{+}\right)=\sqrt{1}\\ =1\end{array}$

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