Chapter 1.5, Problem 15E

(a)

To determine

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

Answer to Problem 15E

  $\left(f+g\right)\left(x\right)=x^2+5+\sqrt{1-x}$

Explanation of Solution

Given: $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

  $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

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

Substituting the values:

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

Conclusion:

Hence, $\left(f+g\right)\left(x\right)=x^2+5+\sqrt{1-x}$

(b)

To determine

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

Answer to Problem 15E

  $\left(f-g\right)\left(x\right)=x^2+5-\sqrt{1-x}$

Explanation of Solution

Given: $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

  $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

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

Substituting the values:

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

Conclusion:

Hence, $\left(f-g\right)\left(x\right)=x^2+5-\sqrt{1-x}$

(c)

To determine

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

Answer to Problem 15E

  $\left(fg\right)\left(x\right)=x^2\sqrt{1-x}+5\sqrt{1-x}$

Explanation of Solution

Given: $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

  $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

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

Substituting the values:

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

Conclusion:

Hence, $\left(fg\right)\left(x\right)=x^2\sqrt{1-x}+5\sqrt{1-x}$

(d)

To determine

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

Answer to Problem 15E

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

Explanation of Solution

Given: $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

  $f\left(x\right)=x^2+5$ and $g\left(x\right)=\sqrt{1-x}$

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

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

Substituting the values:

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

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

Denominator can never be zero and also square root can never be negative.

Combining the above two conditions:

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

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

Conclusion:

Hence, $\left(f/g\right)\left(x\right)=\frac{x^2+5}{\sqrt{1-x}}$ and domain is $\left(-\infty ,1\right)$