Q. 35 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\left\{\begin{array}{l}1+x if x<0\\ x^2 if x\ge 0\end{array}\right.$

(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 of the given function is the set of all the real numbers. 

(b)  The x and y intercepts are $(-1,0) and (0,0)$ respectively

(c) Graph of the function

(d) Range of the function is set of all the real numbers

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

Step 1.Given information

The given function $f\left(x\right)=\left\{\begin{array}{l}1+x if x<0\\ x^2 if x\ge 0\end{array}\right.$

Step 2.Find the domain of each function

 The domain of the function $f\left(x\right)$ , is the set of all the possible values of x.

The value of the function $f\left(x\right)$ when $x<0$ is given by 1+x .
In the expression $1+x,1$ is added to x. This operation can be performed on any real number.

The value of the function $f\left(x\right)$  when $x\ge 1$ is given by x².
In the expression x² , the value of the variable x is raised to the power of 2 . These operations can be performed on any real number.
So, the domain of x is the set of all real numbers.
Therefore, the domain of the given function is the set of all the real numbers.

   

Step 3.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.

Determine the points on the graph for which the x-coordinate is 0.
The value of the function $f\left(x\right)$ or the y-coordinate when $x=0$ is given by x². 

   $$\begin{gathered} f\left(0\right)=0^2 \\ f\left(0\right)=0 \end{gathered}$$

So,the y-intercept is 0.

Determine the points on the graph for which the y -coordinate is 0 .
Consider the function $f\left(x\right)=1+x$ . If the y-coordinate is 0 , then $f\left(x\right)=0$.

$$\begin{gathered} 0=1+x \\ x=-1 \end{gathered}$$

So, the x -intercept is -1 .
Therefore the intercepts are (-1,0) and (0,0)

Step 4.Graph each function

For plotting the graph of the line, $y=1+x$ , in the part for which $x<0$, substitute various values for x  in the equation to obtain some points on the graph.

x	y=1+x	(x,y)
-1	0	(-1,0)
0	1	(0,1)

For plotting the graph of the parabola $y=x^2$,  in the part for which $x\ge 0$ , substitute various values for x  in the equation to obtain some points on the graph. 

x	y=x2	(x,y)
o	0	(0,0)
0.5	0.25	(0.5,0.25)
1	1	(1,1)

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 points plotted on the graph we can see that for each number y , there is at least one number x  in the domain.
So, the range of f  is the set of all the real numbers. 

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:

$$\begin{gathered} f\left(0^{-}\right)=1+x \\ =1+0 \\ = 1 \\ and \\ f\left(0^{+}\right)=x^2 \\ =0^2 \\ =0 \end{gathered}$$

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

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  $x=0$.