Chapter 2: Q. 71 (page 69)

In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.
$(a) (f + g) (x) (b) (f - g) (x) (c) (f \cdot g) (x) (d) (\frac{f}{g}) (x) (e) (f + g) (3) (f) (f - g) (4) (g) (f \cdot g) (2) (h) (\frac{f}{g}) (1)$
$f (x) = \frac{2 x + 3}{3 x - 2}; g (x) = \frac{4 x}{3 x - 2}$

Short Answer

Expert verified

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

The value of role="math" localid="1646147453039" $(f - g) (x) = \frac{3 - 2 x}{3 x - 2}$ and the domain of $f - g$ is ${x | x \neq \frac{2}{3}}$

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

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

The value of $(f + g) (3) = 3$

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

The value of $(f \cdot g) (2) = \frac{7}{2}$

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

Step by step solution

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.

$(a) (f + g) (x) (b) (f - g) (x) (c) (f \cdot g) (x) (d) (\frac{f}{g}) (x) (e) (f + g) (3) (f) (f - g) (4) (g) (f \cdot g) (2) (h) (\frac{f}{g}) (1)$

The given function is $f (x) = \frac{2 x + 3}{3 x - 2}; g (x) = \frac{4 x}{3 x - 2}$

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 $f (x) \neq 0$ and $g (x) \neq 0$

role="math" localid="1646141944459" $3 x - 2 \neq 0 3 x - 2 + 2 \neq 0 + 2 3 x \neq 2 \frac{3}{3} x \neq \frac{2}{3} x \neq \frac{2}{3}$

The domain ofrole="math" localid="1646141988750" $f (x) = {x | x \neq \frac{2}{3}} and g (x) = {x | x \neq \frac{2}{3}}$

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

We know that $(f + g) (x) = f (x) + g (x)$

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

$(f + g) (x) = \frac{2 x + 3}{3 x - 2} + \frac{4 x}{3 x - 2} (f + g) (x) = \frac{2 x + 3 + 4 x}{3 x - 2} (f + g) (x) = \frac{6 x + 3}{3 x - 2} (f + g) (x) = \frac{6 x + 3}{3 x - 2}$

The domain of $f + g$ consists of those numbers x that are in the domains of both fandg. Therefore, the domain of $f + g$ is ${x | x \neq \frac{2}{3}}$

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 $f (x) and g (x)$

role="math" localid="1646142706097" $(f - g) (x) = \frac{2 x + 3}{3 x - 2} - \frac{4 x}{3 x - 2} (f - g) (x) = \frac{2 x + 3 - 4 x}{3 x - 2} (f - g) (x) = \frac{3 - 2 x}{3 x - 2}$

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

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 $f (x) and g (x)$

$(f \cdot g) (x) = \frac{2 x + 3}{3 x - 2} \cdot \frac{4 x}{3 x - 2} (f \cdot g) (x) = \frac{4 x (2 x + 3)}{{(3 x - 2)}^{2}}$

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

$(f \cdot g) (x) = \frac{4 x \cdot 2 x + 4 x \cdot 3}{{(3 x)^{2} - 2 \times 3 x \times 2 + (2)}^{2}} (f \cdot g) (x) = \frac{8 x^{2} + 12 x}{9 x^{2} - 12 x + 4}$

The domain of $f \cdot g$ consists of those numbers x that are in the domains of both $f and g$ . Therefore, the domain of $f \cdot g$ is ${x | x \neq \frac{2}{3}}$ .

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

Putting the value of $f (x) and g (x)$

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

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

Since $g (x) \neq 0$ when $4 x \neq 0$

$\frac{4}{4} x \neq \frac{0}{4} x \neq 0$

The domain of $\frac{f}{g}$ is ${x | x \neq 0}$

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)=6x+33x-2

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

$(f + g) (3) = \frac{6 \cdot 3 + 3}{3 \cdot 3 - 2} (f + g) (3) = \frac{18 + 3}{9 - 2} (f + g) (3) = \frac{21}{7} (f + g) (3) = 3$

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)=3-2x3x-2

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

$(f - g) (4) = \frac{3 - 2 \cdot 4}{3 \cdot 4 - 2} (f - g) (4) = \frac{3 - 8}{12 - 2} (f - g) (4) = \frac{- 5}{10} (f - g) (4) = - \frac{1}{2}$

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)=8x2+12x9x2-12x+4

Putting $x = 2$ in the value of $(f \cdot g) (x)$

$(f \cdot g) (2) = \frac{8 {(2)}^{2} + 12 \cdot 2}{9 {(2)}^{2} - 12 \cdot 2 + 4} (f \cdot g) (2) = \frac{8 \cdot 4 + 24}{9 \cdot 4 - 24 + 4} (f \cdot g) (2) = \frac{32 + 24}{36 - 24 + 4} (f \cdot g) (2) = \frac{56}{16} (f \cdot g) (2) = \frac{7}{2}$

Part (h) Step 1. We have to find the value of fg(1)From the part (a) we know the value of fg(x)=2x+34x

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

$(\frac{f}{g}) (1) = \frac{2 \cdot 1 + 3}{4 \cdot 1} (\frac{f}{g}) (1) = \frac{2 + 3}{4} (\frac{f}{g}) (1) = \frac{5}{4}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Step 1. Given Information

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.

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

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

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

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

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

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

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)=6x+33x-2

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)=3-2x3x-2

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)=8x2+12x9x2-12x+4

Part (h) Step 1. We have to find the value of fg(1)From the part (a) we know the value of fg(x)=2x+34x

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Discrete Mathematics

Theoretical and Mathematical Physics

Probability and Statistics

Logic and Functions

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.(a)(f+g)(x)(b)(f-g)(x)(c)(f·g)(x)(d)fg(x)(e)(f+g)(3)(f)(f-g)(4)(g)(f·g)(2)(h)fg(1)f(x)=2x+33x-2;g(x)=4x3x-2

Short Answer

Step by step solution

Step 1. Given Information

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.

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

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

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

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

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

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

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)=6x+33x-2

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)=3-2x3x-2

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)=8x2+12x9x2-12x+4

Part (h) Step 1. We have to find the value of fg(1)From the part (a) we know the value of fg(x)=2x+34x

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Discrete Mathematics

Theoretical and Mathematical Physics

Probability and Statistics

Logic and Functions

Applied Mathematics

Study anywhere. Anytime. Across all devices.