Chapter 1.2, Problem 65E

(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 65E

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 $\left[-2,\infty \right)$ and $\left[-3,\infty \right)$ respectively. The domain and range of the function $g\circ f$ is $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ and $\left[0,\infty \right)$ respectively.

Explanation of Solution

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

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 $\left(f\circ g\right)\left(x\right)$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{X}}\boxed{^}\boxed{(}\boxed{2}\boxed{)}\boxed{-}\boxed{3}$ . Press Enter after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{2}\boxed{)}\boxed{^}\boxed{(}\boxed{1}\boxed{÷}\boxed{2}\boxed{)}$ . 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,

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

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 greater than $-2$ . So, the domain of the function is the interval $\left[-2,\infty \right)$ .

As observed from the graph of function $\left(f\circ g\right)\left(x\right)$ has the minimum value as $-3$ and no maximum value. So, the range of the function is the interval $\left[-3,\infty \right)$ .

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 $\left(gof\right)\left(x\right)$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{\text{X}}\boxed{^}\boxed{(}\boxed{2}\boxed{)}\boxed{-}\boxed{3}$ . Press Enter after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{2}\boxed{)}\boxed{^}\boxed{(}\boxed{1}\boxed{÷}\boxed{2}\boxed{)}$ . 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,

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

Now press the enter button on the equal button 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 $x$ except between $-1$ and $1$ . So, the domain of the function is the interval $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ .

As observed from the graph of function $\left(g\circ f\right)\left(x\right)$ has the minimum value as $0$ and no maximum value. So, the range of the function is the interval $\left[0,\infty \right)$ .

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 $\left[-2,\infty \right)$ and $\left[-3,\infty \right)$ respectively. The domain and range of the function $g\circ f$ is $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ and $\left[0,\infty \right)$ 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 65E

The conjectures for the domain and range in part (a) are true.

Explanation of Solution

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

Calculation:

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

  $f\left(g\left(x\right)\right)=f\left(\sqrt{x+2}\right)$

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

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

The function $\left(f\circ g\right)\left(x\right)=x-1$ is defined for all the real values of $x$ greater than or equal to $-2$ as $x+2\ge 0$ . So, the domain of the function is the interval $\left[-2,\infty \right)$ .

The function $\left(f\circ g\right)\left(x\right)=x-1$ has the minimum value as $-3$ and no maximum value in the interval $\left[-2,\infty \right)$ . So, the range of the function is the interval $\left[-3,\infty \right)$ .

To find the composite function $g\left(f\left(x\right)\right)$ , substitute $x^2-3$ for $f\left(x\right)$ in the function $g\left(f\left(x\right)\right)$ .

  $g\left(f\left(x\right)\right)=g\left(x^2-3\right)$

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

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

The function $\left(g\circ f\right)\left(x\right)=\sqrt{x^2-1}$ is defined for all the real values of $x$ such that $x^2-1\ge 0$ . So, the domain of the function is the interval $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ .

The function $\left(g\circ f\right)\left(x\right)=\sqrt{x^2-1}$ has the minimum value as $0$ and no maximum value for the interval of domain $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ . So, the range of the function is the interval $\left[0,\infty \right)$ .

Therefore, the conjectures for the domain and range in part (a) is true.