Q. 37 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 29–40:

$f\left(x\right)=\left\{\begin{array}{l}\left|x\right| if -2\le x<0\\ x^3 if x>0\end{array}\right.$

1. Find the domain of each function.
2. Locate any intercepts.
3. Graph each function.
4. Based on the graph, Find the range.
5. Is f continuous on its domain?

Step-by-Step Solution

Answer

1. The domain of the function is $[-2,0)\cup (0,\infty )$.
2. There is no intercepts.
3. The graph of the function is as follows,
   
4. The range of the function is $(0,\infty )$.
5. The function f  is not continuous at $x=0$.

Step 1. Given Information

We are given a function,  

$f\left(x\right)=\left\{\begin{array}{l}\left|x\right| if -2\le x<0\\ x^3 if x>0\end{array}\right.$

Part(a) Step 1. Finding the domain of the function

The value of function $f\left(x\right)=\left|x\right|$ lies between $0$ and $-2$ inclusive.

The function $f\left(x\right)=x^3$ is defined for $x>0$

So, the domain of the function is,

$[-2,0)\cup (0,\infty )$

Part(b) Step 1. Locate any intercepts

The x-intercepts are those points for which the y -coordinate is zero and the y-intercepts are those points for which the x-coordinate is zero.  

The function is not defined for $x=0$,

Hence there is no intercepts.

Part(c) Step 1. Graphing the function

The table of values for graphing the function are as follows,

So, the graph of given function is as follows,

Part(d) Step 1. Finding the range

As seen from the graph, the range of the function is $(0,\infty )$

Part(e) Step 1. Checking continuity

The function f  is not continuous because there is a jump in the graph at $x=0$.