Q. 37 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 37 and 38,

(a) Find the domain of each function. 

(b) Locate any intercepts. 

(c) Graph each function.

(d) Based on the graph, find the range. 

(e) Is f continuous on its domain?

If

$f\left(x\right)=\left\{\begin{array}{ll}3x& if -2<x\le 1\\ x+1& if x>1\end{array}\right.$

Step-by-Step Solution

Answer

Part a. The domain of the function is $\left\{x\right|x>-2\}$or the interval $(-2,\infty ).$

Part b. The intercept is $(0,0).$

Part c. The graph of the function is

Part d. The range of the function is $\left\{y\right|y\ge -6\}$or the interval $(-6,\infty ).$

Part e. No, f  is not continuous on its domain.

Part (a) Step 1. Given Information

The given function is $f\left(x\right)=\left\{\begin{array}{ll}3x& if -2<x\le 1\\ x+1& if x>1\end{array}\right.$

We have to find the domain of each function.

Part (a) Step 2. Finding the domain of the function

The domain of the function is defined as the set of all possible values of x.

In the function $f\left(x\right)=3x$, the operation can be performed between the real numbers $-2 and 1.$

In the function $f\left(x\right)=x+1$, the operation can be performed on any real number greater than $1.$

Thus, the domain is the set of numbers greater than $-2.$

The domain of the given function is $\left\{x\right|x>-2\} or the interval \{-2,\infty ).$

Part (b) Step 1. Given Information

The given function is

$f\left(x\right)=\left\{\begin{array}{ll}3x& if -2<x\le 1\\ x+1& if x>1\end{array}\right.$

We have to locate any intercepts.

Part (b) Step 2. Finding the intercepts

The x-intercepts are the points where the y-coordinate is $0$ and the y-intercepts are the points where x-coordinate is 

Thus, when $x=0$

$$\begin{gathered} f\left(0\right)=3\left(0\right) \\ f\left(0\right)=0 \end{gathered}$$

Then $y=0$

Therefore, intercept is $(0,0).$

Part (c) Step 1. Sketching the graph of the function

To graph the function, we graph each piece.

First, we graph the line $f\left(x\right)=3x$ and keep the only part for which $-2<x\le 1.$

Then we graph the line $f\left(x\right)=x+1$ and keep the only part for which $x>1.$

Part (c) Step 2. Graph of the function

The graph of the function is

Part (d) Step 1. Finding the range

From the graph, we conclude that all the y-coordinate is greater than $-6.$

So, the range of the function is $\left\{y\right|y\ge -6\} or the interval (-6,\infty ).$

Part (e) Step 1. Determining is f continuous in its domain

From the graph, we conclude that the function is not continuous in its domain because there is a "jump" at $x=1.$