Chapter EP, Problem 7.1.2EP

To determine

To calculate: The value of $\left(f+g\right)\left(x\right),\left(f-g\right)\left(x\right),\left(f\cdot g\right)\left(x\right),\text{ and }\left(\frac{f}{g}\right)\left(x\right)$ where $f\left(x\right)=\sqrt{x}\text{ and }g\left(x\right)=x^2$ .

Answer to Problem 7.1.2EP

The value of

  $\left(f+g\right)\left(x\right)=\sqrt{x}+x^2,\left(f-g\right)\left(x\right)=\sqrt{x}-x^2,\left(f\cdot g\right)\left(x\right)=x^2\sqrt{x},\text{ and }\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{x^2}$ , $x\ne 0$ .

Explanation of Solution

Given information:

The functions $f\left(x\right)=\sqrt{x}\text{ and }g\left(x\right)=x^2$ .

Formula used:

The sum of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ .

The difference of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$ .

The product of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$ .

The quotient of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ .

Calculation:

Consider the functions $f\left(x\right)=\sqrt{x}\text{ and }g\left(x\right)=x^2$ .

Recall that sum of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ .

Apply it,

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

Recall that difference of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$ .

Apply it,

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

Recall that product of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$ .

Apply it,

  $\begin{array}{c}\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)\\ =\left(\sqrt{x}\right)\left(x^2\right)\\ =x^2\sqrt{x}\end{array}$

Recall that quotient of two functions $f\left(x\right)\text{and }g\left(x\right)$ is $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ .

Apply it,

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

The domain of the above expression is defined when denominator is not equal to 0.

That is,

  $\begin{array}{c}{x}^2\ne 0\\ x\ne 0\end{array}$

For the quotient to be defined $x\ne 0$ .

Thus, the value of

  $\left(f+g\right)\left(x\right)=\sqrt{x}+x^2,\left(f-g\right)\left(x\right)=\sqrt{x}-x^2,\left(f\cdot g\right)\left(x\right)=x^2\sqrt{x},\text{ and }\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{x^2}$ , $x\ne 0$ .