Q. 67 - 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.

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

$f\left(x\right)=\sqrt{x}; g\left(x\right)=3x-5$

Step-by-Step Solution

Answer

The value of $(f+g)\left(x\right)=\sqrt{x}+3x-5$ and the domain of $f+g$ is $\left\{x\right|x\ge 0\}$

The value of $(f-g)\left(x\right)=\sqrt{x}-3x+5$ and the domain of $f-g$ is $\left\{x\right|x\ge 0\}$

The value of $(f·g)\left(x\right)=3x\sqrt{x}-5\sqrt{x}$ and the domain of $f·g$ is $\left\{x\right|x\ge 0\}$.

The value of $\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{3x-5}$ and the domain of $\frac{f}{g}$ is $\left\{x\right|x\ge 0,x\ne \frac{5}{3}\}$.

The value of $(f+g)\left(3\right)=\sqrt{3}+4$

The value of $(f-g)\left(4\right)=-5$

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

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

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  $f\left(x\right)=\sqrt{x}; g\left(x\right)=3x-5$

Step 2. The function f tells us to take the square root of x . 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 $x\ge 0$ so the domain of $f\left(x\right) \mathrm{is} \left\{\mathrm{x}\right|\mathrm{x}\ge 0\}$.

Function g tells us three times of a number then subtract five. Since these operations can be performed on any real number, we conclude that the domain of  is the set of all real numbers.

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 )

$(f+g)\left(x\right)=\sqrt{x}+3x-5$

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 $\left\{x\right|x\ge 0\}$.

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 $f\left(x\right) \mathrm{and} g\left(x\right)$

$$\begin{gathered} (f-g)\left(x\right)=\sqrt{x}-\left(3x-5\right) \\ (f-g)\left(x\right)=\sqrt{x}-3x+5 \end{gathered}$$

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

$\left\{x\right|x\ge 0\}$.

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 $f\left(x\right) \mathrm{and} g\left(x\right)$

$$\begin{gathered} (f·g)\left(x\right)=\sqrt{x}·(3x-5) \\ (f·g)\left(x\right)=3x\sqrt{x}-5\sqrt{x} \end{gathered}$$

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 $\left\{x\right|x\ge 0\}$.

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 $f\left(x\right) \mathrm{and} g\left(x\right)$

$\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{3x-5}$

Part (d) Step 2. The domain of f g consists of the numbers for which g ( x ) ≠ 0 and that are in the domains of both f   and   g .

This requires that $x\ge 0$

Since $g\left(x\right)\ne 0$ when $3x-5\ne 0$

Add 5 on both side

$$\begin{gathered} 3x-5+5\ne 0+5 \\ 3x\ne 5 \\ \frac{3}{3}x\ne \frac{5}{3} \\ x\ne \frac{5}{3} \end{gathered}$$

The domain of $\frac{f}{g} \mathrm{is}$$\left\{x\right|x\ge 0,x\ne \frac{5}{3}\}$

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 ) = 3 + 3 x - 5

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

$$\begin{gathered} (f+g)\left(3\right)=\sqrt{3}+3·3-5 \\ (f+g)\left(3\right)=\sqrt{3}+9-5 \\ (f+g)\left(3\right)=\sqrt{3}+4 \end{gathered}$$

Part (f) Step 1. We have to find the value of ( f - g ) ( 4 ) From the part (b) we know the value of ( f - g ) ( x ) = x - 3 x + 5

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

$$\begin{gathered} (f-g)\left(4\right)=\sqrt{4}-3·4+5 \\ (f-g)\left(4\right)=2-12+5 \\ (f-g)\left(4\right)=-5 \end{gathered}$$

Part (g) Step 1. We have to find the value of ( f · g ) ( 2 ) From the part (c) we know the value of ( f · g ) ( x ) = 3 x x - 5 x

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

$$\begin{gathered} (f·g)\left(2\right)=3·2\sqrt{2}-5\sqrt{2} \\ (f·g)\left(2\right)=6\sqrt{2}-5\sqrt{2} \\ (f·g)\left(2\right)=\sqrt{2} \end{gathered}$$

Part (h) Step 1. We have to find the value of f g ( 1 ) From the part (d) we know the value of f g ( x ) = x 3 x - 5

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

$$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=\frac{\sqrt{1}}{3·1-5} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{1}{3-5} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{1}{-2} \end{gathered}$$