Q. 32 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\left\{\begin{array}{l}x+3 if x<-2\\ -2x-3 if x\ge 2\end{array}\right.$

(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?

Step-by-Step Solution

Answer

(a) The domain of the function is the set of all real numbers as there is no value of x where the function is not defined. 

(b) The x -intercepts are $(-3,0),(-\frac{3}{2},0)$ and y-intercepts $(0,-3)$

(c) The graph of the given function

(d)  Range of the given function is $\left\{y|y\le 1\right\}$

(e)  The given function is continous on its domain

step 1.Given information

The given function  $f\left(x\right)=\left\{\begin{array}{l}x+3 if x<-2\\ -2x-3 if x\ge 2\end{array}\right.$

Step 2.Find the domain of each function

We will use the following definition of the function: 

$f\left(x\right)=\left\{\begin{array}{l}x+3 if x<-2\\ -2x-3 if x\ge 2\end{array}\right.$

From the above definition we see the domain of the function is the set of all real numbers as there is no value of x where the function is not defined. 

Step 3.Locate any intercepts

The intercepts are the points on the graph which are obtained when it cuts the x and y axes. 

The points $(-3,0),(-\frac{3}{2},0) and (0,-3)$    satisfy the function above.

Hence the x -intercepts are $(-3,0),(-\frac{3}{2},0)$ .
and the y-intercept is $(0,-3)$ .

Step 4.Graph each function

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

Step 5.Based on the graph,Find the range

 From the graph we see that $f\left(x\right)\le 1$ in its domain. So the range of the function $f\left(x\right)$ is the set$\left\{y|y\le 1\right\}$.

Step 6.Checking f is continuous on its domain?

The only point at which the function might have behaved in a manner that it becomes discontinuous is $x=-2$. But at this point, the value of the function from the left of $-2$ and the right of $-2$ are given as:

$$\begin{gathered} f(-2)=-2+3 \\ =1 \\ and \\ f(-2)=-2(-2)-3 \\ =4-3 \\ =1 \end{gathered}$$

Hence,$f\left(2^{-}\right)=f\left(2^{+}\right)$

So even at the break point the function is continuous.
Hence the function is continuous at all points.
Also from the graph it can be clearly seen that the function is continuous throughout its domain.