Q. 72 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.

$$\begin{gathered} \left(a\right) (f+g)\left(x\right) \left(b\right) (f-g)\left(x\right) \left(c\right) (f·g) \left(x\right) \left(d\right) \left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right) (f+g)\left(3\right) \left(f\right) (f-g)\left(4\right) \left(g\right) ( f·g)\left(2\right) \left(h\right) \left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

$f\left(x\right)=\sqrt{x+1}; g\left(x\right)=\frac{2}{x}$

Step-by-Step Solution

Answer

The value of $(f+g)\left(x\right)=\frac{x\sqrt{x+1}+2}{x}$ and the domain of $f+g$ is $\left\{x\right|x\ge -1,x\ne 0\}$.

The value of $(f-g)\left(x\right)=\frac{x\sqrt{x+1}-2}{x}$ and the domain of $f-g$ is $\left\{x\right|x\ge -1,x\ne 0\}$.

The value of $(f·g)\left(x\right)=\frac{2}{x}\sqrt{x+1}$ and the domain of $f·g$ is $\left\{x\right|x\ge -1,x\ne 0\}$.

The value of $\left(\frac{f}{g}\right)\left(x\right)=\frac{x}{2}\sqrt{x+1}$ and the domain of $\frac{f}{g}$ is $\left\{x\right|x\ge -1\}$.

The value of $(f+g)\left(3\right)=\frac{8}{3}$

The value of $(f-g)\left(4\right)=\frac{2(2\sqrt{5}-1)}{4}$

The value of $(f·g)\left(2\right)=\sqrt{3}$

The value of $\left(\frac{f}{g}\right)\left(1\right)=\frac{1}{\sqrt{2}}$

Step 1. Given Information

In the given problems we have to solve the given functions f and g, find the following. For parts (a)–(d), we have to also find the domain.

$$\begin{gathered} \left(a\right) (f+g)\left(x\right) \left(b\right) (f-g)\left(x\right) \left(c\right) (f·g) \left(x\right) \left(d\right) \left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right) (f+g)\left(3\right) \left(f\right) (f-g)\left(4\right) \left(g\right) ( f·g)\left(2\right) \left(h\right) \left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

The given function is $f\left(x\right)=\sqrt{x+1}; g\left(x\right)=\frac{2}{x}$

Step 2. The function f tells us to take the square root of x + 1 . But only non-negative numbers have real square roots, so the expression under the square root (the radicand) must be non-negative (greater than or equal to zero).

This requires that

$$\begin{gathered} x+1\ge 0 \\ x+1-1\ge 0-1 \\ x\ge -1 \end{gathered}$$ 

so the domain of $f\left(x\right) \mathrm{is} \left\{x\right|x\ge -1\}$.

Function g tells us two divided by a number.

This requires that $x\ne 0$

So the domain of $g\left(x\right) \mathrm{is} x\ne 0$

Part (a) Step 1. We have to find the value of ( f + g ) ( x )

We know that $(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)$

Part (a) Step 2. Putting the value of f ( x )   and   g ( x )

$$\begin{gathered} (f+g)\left(x\right)=\sqrt{x+1}+\frac{2}{x} \\ (f+g)\left(x\right)=\sqrt{x+1}·\frac{x}{x}+\frac{2}{x} \\ (f+g)\left(x\right)=\frac{x\sqrt{x+1}+2}{x} \end{gathered}$$

The domain of $f+g$ consists of those numbers x that are in the domains of both f and g. Therefore, the domain of$f+g$is $\left\{x\right|x\ge -1,x\ne 0\}$.

Part (b) Step 1. We have to find the value of ( f - g ) ( x ) We know ( f - g ) ( x ) = f ( x ) - g ( x )

Putting the value of $f \mathrm{and} g$

$$\begin{gathered} (f-g)\left(x\right)=\sqrt{x+1}-\frac{2}{x} \\ (f-g)\left(x\right)=\sqrt{x+1}·\frac{2}{x}-\frac{2}{x} \\ (f-g)\left(x\right)=\frac{x\sqrt{x+1}-2}{x} \end{gathered}$$

The domain of $f-g$ consists of those numbers x that are in the domains of both . Therefore, the domain of $f-g$ is $\left\{x\right|x\ge -1,x\ne 0\}$.

Part (c) Step 1. We have to find the value of ( f · g ) ( x ) We know that ( f · g ) ( x ) = f ( x ) · g ( x )

Putting the value of $f\left(x\right) \mathrm{and} g\left(x\right)$

$$\begin{gathered} (f·g)\left(x\right)=\sqrt{x+1}·\frac{2}{x} \\ (f·g)\left(x\right)=\frac{2}{x}\sqrt{x+1} \end{gathered}$$

The domain of $f·g$ consists of those numbers x that are in the domains of both $f \mathrm{and} g$. Therefore, the domain of $f·g$ is $\left\{x\right|x\ge -1,x\ne 0\}$.

Part (d) Step 1. We have to find the value of f g ( x ) We know that f g ( x ) = f ( x ) g ( x )

Putting the value of $f\left(x\right) \mathrm{and} g\left(x\right)$

$$\begin{gathered} \left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x+1}}{\frac{2}{x}} \\ \left(\frac{f}{g}\right)\left(x\right)=\sqrt{x+1}·\frac{x}{2} \\ \left(\frac{f}{g}\right)\left(x\right)=\frac{x}{2}\sqrt{x+1} \end{gathered}$$

Part (d) Step 2. The domain of f g are in the domains of both f   and   g .

This requires $$\begin{gathered} x+1\ge 0 \\ x\ge -1 \end{gathered}$$

The domain of $\frac{f}{g}$ is $\left\{x\right|x\ge -1\}$

Part (e) Step 1. We have to find the value of ( f + g ) ( 3 ) From the part (a) we know the value of ( f + g ) ( x ) = x x + 1 + 2 x

Putting $x=3$ in the value of $(f+g)\left(x\right)$

$$\begin{gathered} (f+g)\left(3\right)=\frac{3\sqrt{2+1}+2}{3} \\ (f+g)\left(3\right)=\frac{3\sqrt{4}+2}{3} \\ (f+g)\left(3\right)=\frac{3\times 2+2}{3} \\ (f+g)\left(3\right)=\frac{6+2}{3} \\ (f+g)\left(3\right)=\frac{8}{3} \end{gathered}$$

Part (f) Step 1. We have to find the value of ( f - g ) ( 4 ) From the part (a) we know the value of ( f - g ) ( x ) = x x + 1 - 2 x

Putting $x=4$ in the value of $(f-g)\left(x\right)$

$$\begin{gathered} (f-g)\left(4\right)=\frac{4\sqrt{4+1}-2}{4} \\ (f-g)\left(4\right)=\frac{4\sqrt{5}-2}{4} \\ (f-g)\left(4\right)=\frac{2(2\sqrt{5}-1)}{4} \end{gathered}$$

Part (g) Step 1. We have to find the value of ( f · g ) ( 2 ) From the part (a) we know the value of ( f · g ) ( x ) = 2 x x + 1

Putting $x=2$ in the value of $(f·g)\left(x\right)$

$$\begin{gathered} (f·g)\left(2\right)=\frac{2}{2}\sqrt{2+1} \\ (f·g)\left(2\right)=\sqrt{3} \end{gathered}$$

Part (h) Step 1. We have to find the value of f g ( 1 ) From the part (a) we know the value of f g ( x ) = x 2 x + 1

Putting $x=1$ in the value of $\left(\frac{f}{g}\right)\left(x\right)$

$$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=\frac{1}{2}·\sqrt{1+1} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{1}{2}·\sqrt{2} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{1}{\sqrt{2}\sqrt{2}}·\sqrt{2} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{1}{\sqrt{2}} \end{gathered}$$