Q. 31 - Precalculus Enhanced with Graphing Utilities

Question

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

(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 given function does not have x-interecept.the y-intercept is 3.

(c)  Graph of the function

(d) The range of f is $\left\{y|y\ge 1\right\}$ or the interval $[1,\infty )$

(e) The given function is countinous on its domain

Step 1.Given information

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

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 $x<1$ is given by $-2x+3$ .
In the expression $-2x+3,x$ is multiplied by -2 and then 3 is added. These operations can be performed on any real number.
The value of the function for any value of $x\ge 1$  is given by $3x-2$ .
In the expression $3x-2$ , the value of the variable is multiplied by 3 and then 2 is subtracted from it. 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 they -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 $-2x+3$

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

So,the y intercept is 3.

 For finding the x -intercept, substitute 0 for y in $f\left(x\right)=-2x+3$ . 

$0=-2x+3$

Add 2x to both sides of the equation.

$$\begin{gathered} 0+2x=-2x+3+2x \\ 2x=3 \end{gathered}$$

Divide both sides of the equation by 2 .

$$\begin{gathered} \frac{2x}{2}=\frac{3}{2} \\ x=\frac{3}{2} \end{gathered}$$

The function $-2x+3$  is defined for $x<1$. Since $\frac{3}{2}$ is greater than 1, the function does not have any x-intercept. 

Step 4.Graph each function

Graph each piece to graph the function.
For plotting the graph of the line,$y=-2x+3$ , in the part for which $x>1$ , substitute various values for  in the equation to obtain few points on the graph.

x	y=-2x+3	(x,y)
0	3	(0,3)
1	1	(1,1)

In order to plot the graph of the line $3x-2$  in the part for which $x\ge 1$, substitute various values for x  in the equation to obtain few points on the graph.

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

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 greater than 1 , inclusive. For each number y , there is at least one number x in the domain.

So, the range of f is $\left\{y|y\ge 1\right\}$  or the interval $[1,\infty )$ .

Step 6.

The only point at which the function might have behaved in a manner that it becomes discontinuous is $x=1$ . But at this point, the value of the function from the left of 1 and the right of 1 are given as:

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

and 

$$\begin{gathered} f\left(1\right)=3\left(1\right)-2 \\ =3-2 \\ =1 \end{gathered}$$

Hence $f\left(1^{-}\right)=f\left(1^{+}\right)$

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.