Q. 34 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\left\{\begin{array}{l}2x+5 if -3\le x<0\\ -3 if x=0\\ -5x if x>0\end{array}\right.$

(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 is $\left\{x|x\ge 3\right\}$ 

(b)   The x and y intercepts are $\left(-\frac{5}{2},0\right)$ and $(0,-3)$ respectively.

(c)  The graph of the given function

(d)  Range of the given function $\left\{y|y<5\right\}$

(e)   The function is discontinuous in its domain only at the point $x=0$  

step 1.Given information

The given function is $f\left(x\right)=\left\{\begin{array}{l}2x+5 if -3\le x<0\\ -3 if x=0\\ -5x if x>0\end{array}\right.$

Step 2.Find the domain of each function

We will use the defination of the function

$f\left(x\right)=\left\{\begin{array}{l}2x+5 if -3\le x<0\\ -3 if x=0\\ -5x if x>0\end{array}\right.$

From the above definition we see the domain of the function is the set of all real numbers such that $x\ge 3$. Hence the domain is  $\left\{x|x\ge 3\right\}$

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 $(-\frac{5}{2},0)$ and $(0,-3)$ satisfies the function above. Hence the x-intercepts is$(-\frac{5}{2},0)$. 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 $f\left(x\right)\le 5$ in its domain. So the range of the function $f\left(x\right)$  is the set $\left\{y|y<5\right\}$.

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 $x=0$ . But at this point, the value of the function from the left of 0 and the right of 0 are given as:

$$\begin{gathered} f\left(0^{-}\right)=\frac{1}{\infty } \\ =-\infty \\ and \\ f\left(0\right)=\sqrt[3]{0} \\ =0 \\ and \\ f\left(0^{+}\right)=\sqrt[3]{0} \\ =0 \end{gathered}$$

Hence, $$\begin{gathered} f\left(0^{+}\right)\ne f\left(0\right) \\ \ne f\left(0^{-}\right) \end{gathered}$$

So at the break point the function is discontinuous.
Hence the function is discontinuous in its domain only at the point $x=0$ .
Also from the graph it can be clearly seen that the function is discontinuous at the point $x=0$.