Q. 71 - 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. 

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

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

Step-by-Step Solution

Answer

The value of $(f+g)\left(x\right)=\frac{6x+3}{3x-2}$ and the domain of $f+g$ is $\left\{x\right|x\ne \frac{2}{3}\}$

The value of $(f-g)\left(x\right)=\frac{3-2x}{3x-2}$ and the domain of $f-g$ is $\left\{x\right|x\ne \frac{2}{3}\}$

The value of $(f·g)\left(x\right)=\frac{8x^2+12x}{9x^2-12x+4}$ and the domain of $f·g$ is $\left\{x\right|x\ne \frac{2}{3}\}$

The value of $\left(\frac{f}{g}\right)\left(x\right)=\frac{2x+3}{4x}$ and the domain of $\frac{f}{g}$ is $\left\{x\right|x\ne 0\}$.

The value of $(f+g)\left(3\right)=3$

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

The value of $(f·g)\left(2\right)=\frac{7}{2}$

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

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

Step 2. The function f tells us two times of a number and add three then divide three times of a number and subtract by two. The function g tells us four times of a number then divide by three times of a number and subtract by two.

This requires $f\left(x\right)\ne 0$ and $g\left(x\right)\ne 0$

$$\begin{gathered} 3x-2\ne 0 \\ 3x-2+2\ne 0+2 \\ 3x\ne 2 \\ \frac{3}{3}x\ne \frac{2}{3} \\ x\ne \frac{2}{3} \end{gathered}$$

The domain of  $f\left(x\right)=\left\{x\right|x\ne \frac{2}{3}\} \mathrm{and} \mathrm{g}(\mathrm{x})=\{\mathrm{x}|\mathrm{x}\ne \frac{2}{3}\}$

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 )

$$\begin{gathered} (f+g)\left(x\right)=\frac{2x+3}{3x-2}+\frac{4x}{3x-2} \\ (f+g)\left(x\right)=\frac{2x+3+4x}{3x-2} \\ (f+g)\left(x\right)=\frac{6x+3}{3x-2} \\ (f+g)\left(x\right)=\frac{6x+3}{3x-2} \end{gathered}$$

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 $\left\{x\right|x\ne \frac{2}{3}\}$

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

$$\begin{gathered} (f-g)\left(x\right)=\frac{2x+3}{3x-2}-\frac{4x}{3x-2} \\ (f-g)\left(x\right)=\frac{2x+3-4x}{3x-2} \\ (f-g)\left(x\right)=\frac{3-2x}{3x-2} \end{gathered}$$

The domain of $f-g$ consists of those numbers x that are in the domains of both . Therefore, the domain of $f-g$ is $\left\{x\right|x\ne \frac{2}{3}\}$

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

$$\begin{gathered} (f·g)\left(x\right)=\frac{2x+3}{3x-2}·\frac{4x}{3x-2} \\ (f·g)\left(x\right)=\frac{4x\left(2x+3\right)}{{\left(3x-2\right)}^2} \end{gathered}$$

Using the formula ${(a+b)}^2=a^2+2ab+b^2$

$$\begin{gathered} (f·g)\left(x\right)=\frac{4x·2x+4x·3}{{\left(3x{)}^2-2\times 3x\times 2+(2\right)}^2} \\ (f·g)\left(x\right)=\frac{8x^2+12x}{9x^2-12x+4} \end{gathered}$$

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 $\left\{x\right|x\ne \frac{2}{3}\}$.

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

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

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

Since $g\left(x\right)\ne 0$ when $4x\ne 0$

$$\begin{gathered} \frac{4}{4}x\ne \frac{0}{4} \\ x\ne 0 \end{gathered}$$

The domain of $\frac{f}{g}$ is $\left\{x\right|x\ne 0\}$

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 ) = 6 x + 3 3 x - 2

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

$$\begin{gathered} (f+g)\left(3\right)=\frac{6·3+3}{3·3-2} \\ (f+g)\left(3\right)=\frac{18+3}{9-2} \\ (f+g)\left(3\right)=\frac{21}{7} \\ (f+g)\left(3\right)=3 \end{gathered}$$

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 ) = 3 - 2 x 3 x - 2

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

$$\begin{gathered} (f-g)\left(4\right)=\frac{3-2·4}{3·4-2} \\ (f-g)\left(4\right)=\frac{3-8}{12-2} \\ (f-g)\left(4\right)=\frac{-5}{10} \\ (f-g)\left(4\right)=-\frac{1}{2} \end{gathered}$$

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 ) = 8 x 2 + 12 x 9 x 2 - 12 x + 4

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

$$\begin{gathered} (f·g)\left(2\right)=\frac{8{\left(2\right)}^2+12·2}{9{\left(2\right)}^2-12·2+4} \\ (f·g)\left(2\right)=\frac{8·4+24}{9·4-24+4} \\ (f·g)\left(2\right)=\frac{32+24}{36-24+4} \\ (f·g)\left(2\right)=\frac{56}{16} \\ (f·g)\left(2\right)=\frac{7}{2} \end{gathered}$$

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 ) = 2 x + 3 4 x

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

$$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=\frac{2·1+3}{4·1} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{2+3}{4} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{5}{4} \end{gathered}$$