Chapter 1.5, Problem 12E

(a)

To determine

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

Answer to Problem 12E

  $\left(f+g\right)\left(x\right)=x-4$

Explanation of Solution

Given: $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

  $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

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

Substituting the values:

  $\begin{array}{l}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ \left(f+g\right)\left(x\right)=\left(2x-5\right)+\left(1-x\right)\\ \left(f+g\right)\left(x\right)=x-4\end{array}$

Conclusion:

Hence, $\left(f+g\right)\left(x\right)=x-4$

(b)

To determine

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

Answer to Problem 12E

  $\left(f-g\right)\left(x\right)=3x-6$

Explanation of Solution

Given: $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

  $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

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

Substituting the values:

  $\begin{array}{l}\left(f-g\right)\left(x\right)=\left(2x-5\right)-\left(1-x\right)\\ \left(f-g\right)\left(x\right)=2x-5-1+x\\ \left(f-g\right)\left(x\right)=3x-6\end{array}$

Conclusion:

Hence, $\left(f-g\right)\left(x\right)=3x-6$

(c)

To determine

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

Answer to Problem 12E

  $\left(fg\right)\left(x\right)=-2x^2+7x-5$

Explanation of Solution

Given: $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

  $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

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

Substituting the values:

  $\begin{array}{l}\left(fg\right)\left(x\right)=\left(2x-5\right)\left(1-x\right)\\ \left(fg\right)\left(x\right)=2x-2x^2-5+5x\\ \left(fg\right)\left(x\right)=-2x^2+7x-5\end{array}$

Conclusion:

Hence, $\left(fg\right)\left(x\right)=-2x^2+7x-5$

(d)

To determine

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

Answer to Problem 12E

  $\left(f/g\right)\left(x\right)=\frac{2x-5}{1-x}$

Explanation of Solution

Given: $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

  $f\left(x\right)=2x-5$ and $g\left(x\right)=1-x$

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

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

Substituting the values:

  $\left(f/g\right)\left(x\right)=\frac{2x-5}{1-x}$

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

Denominator can never be zero

  $\begin{array}{l}1-x\ne 0\\ x\ne 1\end{array}$

Thus, domain is $\left(-\infty ,\infty \right)-\left\{1\right\}$

Conclusion:

Hence, $\left(f/g\right)\left(x\right)=\frac{2x-5}{1-x}$ and domain is all set of real numbers except $1$