Chapter 1.2, Problem 66E

(a)

To determine

To find: The graph of functions $f\circ g$ , $g\circ f$ . Also, write the domain and range of both the functions.

Answer to Problem 66E

The graph of $f\circ g$ , $g\circ f$ are shown in Figure (1) and Figure(2) respectively. The domain and range of the function $f\circ g$ is $\left(-\infty ,\infty \right)$ and $\left(-\infty ,\infty \right)$ respectively. The domain and range of the function $g\circ f$ is $\left(-\infty ,\infty \right)$ and $\left(-\infty ,\infty \right)$ respectively.

Explanation of Solution

Given information: The functions are $f\left(x\right)=\frac{2x-1}{x+3}$ and $g\left(x\right)=\frac{3x+1}{2-x}$ .

Calculation:

To graph a function $y=\left(f\circ g\right)\left(x\right)$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $\left(fog\right)\left(x\right)$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{2}\boxed{\times }\boxed{\text{X}}\boxed{-}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{3}\boxed{)}$ . Press Enter after the symbol ${\text{Y}}_2$ and enter the keystrokes $\boxed{(}\boxed{3}\boxed{\times }\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{2}\boxed{-}\boxed{\text{X}}\boxed{)}$ . Now press Enter after the symbol ${\text{Y}}_3$ . Enter the keystrokes $\boxed{\text{VARS}}\boxed{\to }\boxed{1}\boxed{1}\boxed{(}\boxed{\text{VARS}}\boxed{\to }\boxed{1}\boxed{2}\boxed{)}$ .

The display will show the equation,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\left(\text{2}*\text{X}-1\right)/\left(\text{X}+3\right)\\ {\text{Y}}_2=\left(3*\text{X}+1\right)/\left(2-\text{X}\right)\\ {\text{Y}}_3={\text{Y}}_1\left({\text{Y}}_2\right)\end{array}$

Press the enter button on the equal sign for the first two equations.

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below:

  

Figure (1)

As observed from the graph of function $\left(f\circ g\right)\left(x\right)$ is defined for all the real values. So, the domain of the function is the interval $\left(-\infty ,\infty \right)$ .

As observed from the graph of function $\left(f\circ g\right)\left(x\right)$ has no minimum value and no maximum value. So, the range of the function is the interval $\left(-\infty ,\infty \right)$ .

To graph a function $y=\left(g\circ f\right)\left(x\right)$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $\left(g\circ f\right)\left(x\right)$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{2}\boxed{\times }\boxed{\text{X}}\boxed{-}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{3}\boxed{)}$ . Press Enter after the symbol ${\text{Y}}_2$ and enter the keystrokes $\boxed{(}\boxed{3}\boxed{\times }\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{2}\boxed{-}\boxed{\text{X}}\boxed{)}$ . Now press Enter after the symbol ${\text{Y}}_3$ . Enter the keystrokes $\boxed{\text{VARS}}\boxed{\to }\boxed{1}\boxed{2}\boxed{(}\boxed{\text{VARS}}\boxed{\to }\boxed{1}\boxed{1}\boxed{)}$ .

The display will show the equation,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\left(\text{2}*\text{X}-1\right)/\left(\text{X}+3\right)\\ {\text{Y}}_2=\left(3*\text{X}+1\right)/\left(2-\text{X}\right)\\ {\text{Y}}_3={\text{Y}}_2\left({\text{Y}}_1\right)\end{array}$

Press the enter button on the equal sign for the first two equations. Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below:

  

Figure (2)

As observed from the graph of function $\left(g\circ f\right)\left(x\right)$ is defined for all the real values of $x$ . So, the domain of the function is the interval $\left(-\infty ,\infty \right)$ .

As observed from the graph of function $\left(g\circ f\right)\left(x\right)$ has the no minimum value and no maximum value. So, the range of the function is the interval $\left(-\infty ,\infty \right)$ .

Therefore, the graph of $f\circ g$ , $g\circ f$ are shown in Figure (1) and Figure(2) respectively. The domain and range of the function $f\circ g$ is $\left(-\infty ,\infty \right)$ and $\left(-\infty ,\infty \right)$ respectively. The domain and range of the function $g\circ f$ is $\left(-\infty ,\infty \right)$ and $\left(-\infty ,\infty \right)$ respectively.

(b)

To determine

To check: Whether the domain and range conjectures in part (a) are true or not with the help of formula for $f\circ g$ , and $g\circ f$ .

Answer to Problem 66E

The conjectures for the domain and range in part (a) is not true.

Explanation of Solution

Given information: The functions are $f\left(x\right)=\frac{2x-1}{x+3}$ and $g\left(x\right)=\frac{3x+1}{2-x}$ .

Calculation:

To find the composite function $f\left(g\left(x\right)\right)$ , substitute $\frac{3x+1}{2-x}$ for $g\left(x\right)$ in the function $f\left(g\left(x\right)\right)$ .

  $f\left(g\left(x\right)\right)=f\left(\frac{3x+1}{2-x}\right)$

As given $f\left(x\right)=\frac{2x-1}{x+3}$ , substitute $\frac{3x+1}{2-x}$ for $x$ in the function.

  $\begin{array}{c}f\left(\frac{3x+1}{2-x}\right)=\frac{2\left(\frac{3x+1}{2-x}\right)-1}{\left(\frac{3x+1}{2-x}\right)+3}\\ =\frac{\frac{6x+2-2+x}{2-x}}{\frac{3x+1+6-3x}{2-x}}\\ =\frac{7x}{7}\\ =x\end{array}$

The function $\left(f\circ g\right)\left(x\right)=x$ is defined for all the real values of $x$ except the value $x=2$ as the denominator can’t be negative. So, the domain of the function is the interval $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .

The function $\left(f\circ g\right)\left(x\right)=x$ has the all the real values as output except the value $2$ as $x\ne 2$ . So, the range of the function is the interval $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .

To find the composite function $g\left(f\left(x\right)\right)$ , substitute $\frac{2x-1}{x+3}$ for $f\left(x\right)$ in the function $g\left(f\left(x\right)\right)$ .

  $g\left(f\left(x\right)\right)=g\left(\frac{2x-1}{x+3}\right)$

As given $g\left(x\right)=\frac{3x+1}{2-x}$ , substitute $\frac{2x-1}{x+3}$ for $x$ in the function.

  $\begin{array}{c}g\left(\frac{2x-1}{x+3}\right)=\frac{3\left(\frac{2x-1}{x+3}\right)+1}{2-\left(\frac{2x-1}{x+3}\right)}\\ =\frac{\frac{6x-3+x+3}{x+3}}{\frac{2x+6-2x+1}{x+3}}\\ =\frac{7x}{7}\\ =x\end{array}$

The function $\left(g\circ f\right)\left(x\right)=x$ is defined for all the real values of $x$ except $x=-3$ . So, the domain of the function is the interval $\left(-\infty ,-3\right)\cup \left(-3,\infty \right)$ .

The function $\left(g\circ f\right)\left(x\right)=x$ has the all the real values as output except the value $-3$ as $x\ne -3$ . So, the range of the function is the interval $\left(-\infty ,-3\right)\cup \left(-3,\infty \right)$ .

Therefore, the conjectures for the domain and range in part (a) is not true.