Chapter 1, Problem 86RE

a.

To determine

Addition of two functions $\left(f+g\right)x$

Answer to Problem 86RE

Addition of the function will be $x^2-4+\sqrt{3-x}$

Explanation of Solution

Given:

Function is given as;

  $f\left(x\right)=x^2-4,g\left(x\right)=\sqrt{3-x}$

General rule of addition of two function is;

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

Therefore

  $\left(f+g\right)x=x^2-4+\sqrt{3-x}$

b.

To determine

Subtraction of two functions $\left(f-g\right)x$

Answer to Problem 86RE

Subtraction of the function will be $x^2-4-\sqrt{3-x}$

Explanation of Solution

Given:

Function is given as;

  $f\left(x\right)=x^2-4,g\left(x\right)=\sqrt{3-x}$

General rule of subtraction of two function is;

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

Therefore

  $\left(f+g\right)x=x^2-4-\sqrt{3-x}$

c.

To determine

Multiplication of two functions $fg\left(x\right)$

Answer to Problem 86RE

Multiplication of the functions will be $x^2\left(\sqrt{3-x}\right)-4\left(\sqrt{3-x}\right)$

Explanation of Solution

Given:

Function is given as;

  $f\left(x\right)=x^2-4,g\left(x\right)=\sqrt{3-x}$

General rule of multiplication of two function is;

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

Therefore

  $\begin{array}{l}fg\left(x\right)=\left(x^2-4\right).\left(\sqrt{3-x}\right)\\ =x^2\left(\sqrt{3-x}\right)-4\left(\sqrt{3-x}\right)\end{array}$

d.

To determine

Division of two function $\frac{f}{g}\left(x\right)$ and domain of it.

Answer to Problem 86RE

Division of functions will be $\frac{x^2-4}{\sqrt{3-x}}$ and domain will be $\left(-\infty ,3\right)$

Explanation of Solution

Given:

Function is given as;

  $f\left(x\right)=x^2-4,g\left(x\right)=\sqrt{3-x}$

General rule of division of two function is;

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

Therefore

  $\frac{f}{g}\left(x\right)=\frac{x^2-4}{\sqrt{3-x}}$

From the denominator it is clear that value of x can not be 3 and will be not be greater than 3

Therefore domain of the $\frac{f}{g}\left(x\right)$ will be $\left(-\infty ,3\right)$