Chapter 1.5, Problem 13E

(a)

To determine

To calculate: To find $\left(f+g\right)\left(x\right)$

Answer to Problem 13E

  $\left(f+g\right)\left(x\right)=3x^2-5x+6$

Explanation of Solution

Given: $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

Formula Used: Addition of two functions $f$ and $g$ is defined as

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

Calculation:

Functions are given as

  $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

  $\left(f+g\right)\left(x\right)$ is calculated as follows:

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

Substituting the values:

  $\begin{array}{l}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ \left(f+g\right)\left(x\right)=3x^2+\left(6-5x\right)\\ \left(f+g\right)\left(x\right)=3x^2-5x+6\end{array}$

Conclusion:

Hence, $\left(f+g\right)\left(x\right)=3x^2-5x+6$

(b)

To determine

To calculate: To find $\left(f-g\right)\left(x\right)$

Answer to Problem 13E

  $\left(f-g\right)\left(x\right)=3x^2+5x-6$

Explanation of Solution

Given: $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

Formula Used: Subtraction of two functions $f$ and $g$ is defined as

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

Calculation:

Functions are given as

  $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

  $\left(f-g\right)\left(x\right)$ is calculated as follows:

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

Substituting the values:

  $\begin{array}{l}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ \left(f-g\right)\left(x\right)=3x^2-\left(6-5x\right)\\ \left(f-g\right)\left(x\right)=3x^2+5x-6\end{array}$

Conclusion:

Hence, $\left(f-g\right)\left(x\right)=3x^2+5x-6$

(c)

To determine

To calculate: To find $\left(fg\right)\left(x\right)$

Answer to Problem 13E

  $\left(fg\right)\left(x\right)=18x^2-15x^3$

Explanation of Solution

Given: $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

Formula Used: Product of two functions $f$ and $g$ is defined as

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

Calculation:

Functions are given as

  $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

  $\left(fg\right)\left(x\right)$ is calculated as follows:

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

Substituting the values:

  $\begin{array}{l}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ \left(fg\right)\left(x\right)=3x^2\left(6-5x\right)\\ \left(fg\right)\left(x\right)=18x^2-15x^3\end{array}$

Conclusion:

Hence, $\left(fg\right)\left(x\right)=18x^2-15x^3$

(d)

To determine

To calculate: To find $\left(f/g\right)\left(x\right)$

Answer to Problem 13E

  $\left(f/g\right)\left(x\right)=\frac{3x^2}{6-5x}$

Explanation of Solution

Given: $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

Formula Used: Quotient of two functions $f$ and $g$ is defined as

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

Calculation:

Functions are given as

  $f\left(x\right)=3x^2$ and $g\left(x\right)=6-5x$

  $\left(f/g\right)\left(x\right)$ is calculated as follows:

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

Substituting the values:

  $\left(f/g\right)\left(x\right)=\frac{3x^2}{6-5x}$

To calculate domain of $\left(f/g\right)\left(x\right)$ :

Denominator can never be zero

  $\begin{array}{l}6-5x\ne 0\\ x\ne \frac{6}{5}\end{array}$

Thus, domain is $\left(-\infty ,\infty \right)-\left\{\frac{6}{5}\right\}$

Conclusion:

Hence, $\left(f/g\right)\left(x\right)=\frac{3x^2}{6-5x}$ and domain is all set of real numbers except $\frac{6}{5}$