Chapter 1.2, Problem 53E

To determine

To find: The missing values in the given table.

Answer to Problem 53E

The complete table is shown below:

	$g\left(x\right)$	$f\left(x\right)$	$\left(f\circ g\right)\left(x\right)$
$\left(a\right)$	$x^2$	$\sqrt{x-5}$	$\sqrt{x^2-5}$
$\left(b\right)$	$\frac{1}{x-1}$	$1+\frac{1}{x}$	$x,x\ne -1$
$\left(c\right)$	$\frac{1}{x}$	$\frac{1}{x}$	$x,x\ne 0$
$\left(d\right)$	$\sqrt{x}$	$x^2$	$\left|x\right|,x\ge 0$

Explanation of Solution

Given information: The given following missing table is:

	$g\left(x\right)$	$f\left(x\right)$	$\left(f\circ g\right)\left(x\right)$
$\left(a\right)$	$?$	$\sqrt{x-5}$	$\sqrt{x^2-5}$
$\left(b\right)$	$?$	$1+\frac{1}{x}$	$x$
$\left(c\right)$	$\frac{1}{x}$	$?$	$x$
$\left(d\right)$	$\sqrt{x}$	$?$	$\left|x\right|,x\ge 0$

Calculation:

(a) The composite function $\left(f\circ g\right)\left(x\right)$ can be written as:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute $g\left(x\right)$ for $x$ in the function $f\left(x\right)=\sqrt{x-5}$ .

  $f\left(g\left(x\right)\right)=\sqrt{g\left(x\right)-5}$

As it is given that $\left(f\circ g\right)\left(x\right)=\sqrt{x^2-5}$ .

Equate both the functions.

  $\begin{array}{c}\sqrt{g\left(x\right)-5}=\sqrt{x^2-5}\\ {\left(\sqrt{g\left(x\right)-5}\right)}^2={\left(\sqrt{x^2-5}\right)}^2\\ g\left(x\right)-5=x^2-5\\ g\left(x\right)=x^2\end{array}$

So, the function $g\left(x\right)$ is equal to $x^2$ .

(b) The composite function $\left(f\circ g\right)\left(x\right)$ can be written as:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

It is given that $f\left(x\right)=1+\frac{1}{x}$ , substitute $g\left(x\right)$ for $x$ in the function.

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

Also, given $\left(f\circ g\right)\left(x\right)=x$ .

Equate both the functions.

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

So, the function $g\left(x\right)$ is equal to $\frac{1}{x-1}$ .

(c) The composite function $\left(f\circ g\right)\left(x\right)$ can be written as:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute $\frac{1}{x}$ for $g\left(x\right)$ and $x$ for $\left(f\circ g\right)\left(x\right)$ in the above function.

  $\begin{array}{c}x=f\left(\frac{1}{x}\right)\\ \frac{1}{\frac{1}{x}}=f\left(\frac{1}{x}\right)\end{array}$

Replace $\frac{1}{x}$ by $x$ in the above function.

  $f\left(x\right)=\frac{1}{x}$

So, the function $f\left(x\right)$ is equal to $\frac{1}{x}$ .

(d) The composite function $\left(f\circ g\right)\left(x\right)$ can be written as:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute $\left|x\right|$ for $\left(f\circ g\right)\left(x\right)$ and $\sqrt{x}$ for $g\left(x\right)$ in the above function.

  $\begin{array}{c}\left|x\right|=f\left(\sqrt{x}\right)\\ \left|\sqrt{x^2}\right|=f\left(\sqrt{x}\right)\end{array}$

Replace $\sqrt{x}$ by $x$ in the above equation.

  $f\left(x\right)=x^2$

So, the function $f\left(x\right)$ is equal to $x^2$ .

Therefore, the complete table is shown below:

	$g\left(x\right)$	$f\left(x\right)$	$\left(f\circ g\right)\left(x\right)$
$\left(a\right)$	$x^2$	$\sqrt{x-5}$	$\sqrt{x^2-5}$
$\left(b\right)$	$\frac{1}{x-1}$	$1+\frac{1}{x}$	$x,x\ne -1$
$\left(c\right)$	$\frac{1}{x}$	$\frac{1}{x}$	$x,x\ne 0$
$\left(d\right)$	$\sqrt{x}$	$x^2$	$\left|x\right|,x\ge 0$