Q. 33 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\left\{\begin{array}{l}x+3 if -2\le x<1\\ 5 if x=1\\ -x+2 if x>1\end{array}\right.$

(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 real numbers.

(b)  The x-intercept and y-intercepts are $(2,0) and (0,3)$ respectively.

(c)  The graph of the function

(d)  The range of the given function is $\left\{y|y<4,y=5\right\}$ or the interval $(-\infty ,4)\cup \left\{5\right\}$

(e)  The given function is discontinuous on its domain

Step 1.Given information

The given function $f\left(x\right)=\left\{\begin{array}{l}x+3 if -2\le x<1\\ 5 if x=1\\ -x+2 if x>1\end{array}\right.$

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 $-2\le x<1$ is given by $x+3$ .

In the expression $x+3,3$ is added to 3 . These operations can be performed on any real number.
The value of the function or the y-coordinate when $x=1$ is 5 .
The value of the function $f\left(x\right)$ when $x>1$ is given by  $-x+2$

In the expression $x+2$, the value of the opposite of the variable x is added to 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 $x+3$ .

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

So, the y-intercept is 3 

Determine the points on the graph for which the y-coordinate is  0.

Consider the function $f\left(x\right)=-x+2$. If the y-coordinate is 0 , then $f\left(x\right)=0$ .

$$\begin{gathered} 0=-x+2 \\ x=2 \end{gathered}$$

So,the x-intercept is $(2,0)$

Therefore, the intercepts are $(0,3)$ and  $(2,0)$

Step 4.Graph each function

For plotting the graph of the line,$y=x+3$ , in the part for which $-2\le x<1$, substitute various values for x  in the equation to obtain some points on the graph. 

x	y=x+3	(x,y)
-2	1	(-2,1)
1	4	(1,4)

The point (1,5)  is also on the graph because, when $x=1,f\left(x\right)=5.$

For plotting the graph of the line, $y=-x+2$, in the part for which $x>1$ , substitute various values for x  in the equation to obtain some points on the graph.

x	y= -x+2	(x,y)
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

The points on the graph of f all have the y-coordinates less than 4 , inclusive. For each number y, there is at least one number x in the domain. $(1,5)$ is a point on the graph. So, the range of f is $\left\{y|y<4,y=5\right\}$ or the interval  $(-\infty ,4)\cup \left\{5\right\}$

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:

$$\begin{gathered} f\left(0^{-}\right)=0+3 \\ =3 \\ and \\ f\left(0\right)=5 \\ and \\ f\left(0^{+}\right)=-x+2 \\ = 0+2 \\ =2 \end{gathered}$$

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 $x=0$.