Chapter 1.8, Problem 8E

a.

To determine

ToEvaluate: $\left(f+g\right)\left(x\right)$ for functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

Answer to Problem 8E

  $\left(f+g\right)\left(x\right)=x^2+3x-15$

Explanation of Solution

Given:

The functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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 $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\left(3x+1\right)+\left(x^2-16\right)\\ =3x+1+x^2-16\\ =x^2+3x-15\end{array}$

Therefore, $\left(f+g\right)\left(x\right)=x^2+3x-15$

Conclusion:

  $\left(f+g\right)\left(x\right)=x^2+3x-15$

b.

To determine

ToEvaluate: $\left(f-g\right)\left(x\right)$ for functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

Answer to Problem 8E

  $\left(f-g\right)\left(x\right)=-x^2+3x+17$

Explanation of Solution

Given:

The functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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 $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =\left(3x+1\right)-\left(x^2-16\right)\\ =3x+1-x^2+16\\ =-x^2+3x+17\end{array}$

Therefore, $\left(f-g\right)\left(x\right)=-x^2+3x+17$

Conclusion:

The $\left(f-g\right)\left(x\right)=-x^2+3x+17$

c.

To determine

ToEvaluate: $\left(fg\right)\left(x\right)$ for functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

Answer to Problem 8E

  $\left(fg\right)\left(x\right)=3x^3+x^2-48x-16$

Explanation of Solution

Given:

The functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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 $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\left(3x+1\right)\left(x^2-16\right)\\ =3x^3-48x+x^2-16\\ =3x^3+x^2-48x-16\end{array}$

Therefore, $\left(fg\right)\left(x\right)=3x^3+x^2-48x-16$

Conclusion:

The $\left(fg\right)\left(x\right)=3x^3+x^2-48x-16$

d.

To determine

ToEvaluate: $\left(f/g\right)\left(x\right)$ for functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$ . And the domain of $f/g$

Answer to Problem 8E

  $\left(f/g\right)\left(x\right)=\frac{3x+1}{x^2-16}$

Thedomain of $\left(f/g\right)\left(x\right)=\frac{3x+1}{x^2-16}$ is $ℝ\backslash \left\{-4,4\right\}$

Explanation of Solution

Given:

The functions $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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 $f\left(x\right)=3x+1$ and $g\left(x\right)=x^2-16$

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

  $\begin{array}{c}\left(f/g\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{3x+1}{x^2-16}\end{array}$

Therefore, $\left(f/g\right)\left(x\right)=\frac{3x+1}{x^2-16}$

The function $\left(f/g\right)\left(x\right)=\frac{3x+1}{x^2-16}$ is not defined for denominator to be zero

That is for $x^2-16=0$

That is $x^2=16$

That is $x=-4$ and $x=4$

Therefore, domain of $\left(f/g\right)\left(x\right)=\frac{3x+1}{x^2-16}$ is $ℝ\backslash \left\{-4,4\right\}$

Conclusion:

The $\left(f/g\right)\left(x\right)=\frac{3x+1}{x^2-16}$

Thedomain of $\left(f/g\right)\left(x\right)=\frac{3x+1}{x^2-16}$ is $ℝ\backslash \left\{-4,4\right\}$