Chapter 1.8, Problem 10E

a.

To determine

ToEvaluate: $\left(f+g\right)\left(x\right)$ for functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

Answer to Problem 10E

  $\left(f+g\right)\left(x\right)=\sqrt{x^2-4}+\frac{x^2}{x^2+1}$

Explanation of Solution

Given:

The functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

Concept Used:

For the functions $f\left(x\right)$ , $g\left(x\right)$

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

Calculation:

Given the functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

The $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\left(\sqrt{x^2-4}\right)+\left(\frac{x^2}{x^2+1}\right)\\ =\sqrt{x^2-4}+\frac{x^2}{x^2+1}\end{array}$

Therefore, $\left(f+g\right)\left(x\right)=\sqrt{x^2-4}+\frac{x^2}{x^2+1}$

Conclusion:

  $\left(f+g\right)\left(x\right)=\sqrt{x^2-4}+\frac{x^2}{x^2+1}$

b.

To determine

ToEvaluate: $\left(f-g\right)\left(x\right)$ for functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

Answer to Problem 10E

  $\left(f-g\right)\left(x\right)=\sqrt{x^2-4}-\frac{x^2}{x^2+1}$

Explanation of Solution

Given:

The functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

Concept Used:

For the functions $f\left(x\right)$ , $g\left(x\right)$

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

Calculation:

Given the functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

The $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =\left(\sqrt{x^2-4}\right)-\left(\frac{x^2}{x^2+1}\right)\\ =\sqrt{x^2-4}-\frac{x^2}{x^2+1}\end{array}$

Therefore, $\left(f-g\right)\left(x\right)=\sqrt{x^2-4}-\frac{x^2}{x^2+1}$

Conclusion:

The $\left(f-g\right)\left(x\right)=\sqrt{x^2-4}-\frac{x^2}{x^2+1}$

c.

To determine

ToEvaluate: $\left(fg\right)\left(x\right)$ for functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

Answer to Problem 10E

  $\left(fg\right)\left(x\right)=\frac{x^2\sqrt{x^2-4}}{x^2+1}$

Explanation of Solution

Given:

The functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

Concept Used:

For the functions $f\left(x\right)$ , $g\left(x\right)$

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

Calculation:

Given the functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

The $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\left(\sqrt{x^2-4}\right)\left(\frac{x^2}{x^2+1}\right)\\ =\frac{x^2\sqrt{x^2-4}}{x^2+1}\end{array}$

Therefore, $\left(fg\right)\left(x\right)=\frac{x^2\sqrt{x^2-4}}{x^2+1}$

Conclusion:

The $\left(fg\right)\left(x\right)=\frac{x^2\sqrt{x^2-4}}{x^2+1}$

d.

To determine

ToEvaluate: $\left(f/g\right)\left(x\right)$ for functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$ . And the domain of $f/g$

Answer to Problem 10E

  $\left(f/g\right)\left(x\right)=\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}$

Thedomain of $\left(f/g\right)\left(x\right)=\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}$ is $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$

Explanation of Solution

Given:

The functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

Concept Used:

For the functions $f\left(x\right)$ , $g\left(x\right)$

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

Calculation:

Given the functions $f\left(x\right)=\sqrt{x^2-4}$ and $g\left(x\right)=\frac{x^2}{x^2+1}$

The $\left(f/g\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

  $\begin{array}{c}\left(f/g\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{\sqrt{x^2-4}}{\frac{x^2}{x^2+1}}\\ =\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}\end{array}$

Therefore, $\left(f/g\right)\left(x\right)=\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}$

The function $\left(f/g\right)\left(x\right)=\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}$ is not defined for denominator to be zero

That is for $x^2=0$

  $x=0$

Also, $x^2-4\ge 0$

That is $x^2\ge 4$

That is $x\ge 2$ and $x\le -2$

Therefore, domain of $\left(f/g\right)\left(x\right)=\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}$ is $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$

Conclusion:

The $\left(f/g\right)\left(x\right)=\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}$

Thedomain of $\left(f/g\right)\left(x\right)=\frac{\left(x^2+1\right)\sqrt{x^2-4}}{x^2}$ is $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$