Math / Algebra

QuestionIn Exercises $15-22$ , find $f(g(x))$ and $g(f(x))$ . State the domain of each.
15. $f(x)=3 x+2 ; g(x)=x-1$
16. $f(x)=x^{2}-1 ; g(x)=\frac{1}{x-1}$
17. $f(x)=x^{2}-2 ; g(x)=\sqrt{x+1}$
18. $f(x)=\frac{1}{x-1} ; g(x)=\sqrt{x}$
19. $f(x)=x^{2} ; g(x)=\sqrt{1-x^{2}}$
20. $f(x)=x^{3} ; g(x)=\sqrt[3]{1-x^{3}}$
21. $f(x)=\frac{1}{2 x} ; g(x)=\frac{1}{3 x}$
22. $f(x)=\frac{1}{x+1} ; g(x)=\frac{1}{x-1}$

Studdy Solution

STEP 1

1. We are given pairs of functions $f(x)$ and $g(x)$ for each exercise.
2. We need to find the compositions $f(g(x))$ and $g(f(x))$ .
3. We need to determine the domain of each composition function.

STEP 2

1. For each exercise, find $f(g(x))$ .
2. Determine the domain of $f(g(x))$ .
3. For each exercise, find $g(f(x))$ .
4. Determine the domain of $g(f(x))$ .
Let's go through each exercise one by one.
Exercise 15:

STEP 3

Given $f(x) = 3x + 2$ and $g(x) = x - 1$ , find $f(g(x))$ :
$f(g(x)) = f(x-1) = 3(x-1) + 2 = 3x - 3 + 2 = 3x - 1$

STEP 4

Determine the domain of $f(g(x)) = 3x - 1$ :
The function $3x - 1$ is a linear function, so its domain is all real numbers, $\mathbb{R}$ .

STEP 5

Find $g(f(x))$ :
$g(f(x)) = g(3x + 2) = (3x + 2) - 1 = 3x + 1$

STEP 6

Determine the domain of $g(f(x)) = 3x + 1$ :
The function $3x + 1$ is a linear function, so its domain is all real numbers, $\mathbb{R}$ .
Exercise 16:

STEP 7

Given $f(x) = x^2 - 1$ and $g(x) = \frac{1}{x-1}$ , find $f(g(x))$ :
$f(g(x)) = f\left(\frac{1}{x-1}\right) = \left(\frac{1}{x-1}\right)^2 - 1$

STEP 8

Determine the domain of $f(g(x)) = \left(\frac{1}{x-1}\right)^2 - 1$ :
The expression $\frac{1}{x-1}$ is undefined when $x = 1$ . Therefore, the domain of $f(g(x))$ is all real numbers except $x = 1$ .

STEP 9

Find $g(f(x))$ :
$g(f(x)) = g(x^2 - 1) = \frac{1}{x^2 - 1 - 1} = \frac{1}{x^2 - 2}$

STEP 10

Determine the domain of $g(f(x)) = \frac{1}{x^2 - 2}$ :
The expression $\frac{1}{x^2 - 2}$ is undefined when $x^2 - 2 = 0$ , i.e., $x^2 = 2$ , which gives $x = \pm\sqrt{2}$ . Therefore, the domain of $g(f(x))$ is all real numbers except $x = \pm\sqrt{2}$ .
Exercise 17:

STEP 11

Given $f(x) = x^2 - 2$ and $g(x) = \sqrt{x+1}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt{x+1}) = (\sqrt{x+1})^2 - 2 = x + 1 - 2 = x - 1$

STEP 12

Determine the domain of $f(g(x)) = x - 1$ :
The function $x - 1$ is a linear function, so its domain is all real numbers, $\mathbb{R}$ . However, $g(x) = \sqrt{x+1}$ requires $x+1 \geq 0$ , i.e., $x \geq -1$ . Thus, the domain of $f(g(x))$ is $x \geq -1$ .

STEP 13

Find $g(f(x))$ :
$g(f(x)) = g(x^2 - 2) = \sqrt{x^2 - 2 + 1} = \sqrt{x^2 - 1}$

STEP 14

Determine the domain of $g(f(x)) = \sqrt{x^2 - 1}$ :
The expression $\sqrt{x^2 - 1}$ is defined when $x^2 - 1 \geq 0$ , i.e., $x^2 \geq 1$ , which gives $x \leq -1$ or $x \geq 1$ . Therefore, the domain of $g(f(x))$ is $(-\infty, -1] \cup [1, \infty)$ .
Exercise 18:

STEP 15

Given $f(x) = \frac{1}{x-1}$ and $g(x) = \sqrt{x}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt{x}) = \frac{1}{\sqrt{x} - 1}$

STEP 16

Determine the domain of $f(g(x)) = \frac{1}{\sqrt{x} - 1}$ :
The expression $\frac{1}{\sqrt{x} - 1}$ is undefined when $\sqrt{x} - 1 = 0$ , i.e., $\sqrt{x} = 1$ , which gives $x = 1$ . Additionally, $g(x) = \sqrt{x}$ requires $x \geq 0$ . Therefore, the domain of $f(g(x))$ is $x \geq 0$ and $x \neq 1$ .

STEP 17

Find $g(f(x))$ :
$g(f(x)) = g\left(\frac{1}{x-1}\right) = \sqrt{\frac{1}{x-1}}$

STEP 18

Determine the domain of $g(f(x)) = \sqrt{\frac{1}{x-1}}$ :
The expression $\sqrt{\frac{1}{x-1}}$ is defined when $\frac{1}{x-1} \geq 0$ , i.e., $x-1 > 0$ (since $\frac{1}{x-1} = 0$ is not possible), which gives $x > 1$ . Therefore, the domain of $g(f(x))$ is $x > 1$ .
Exercise 19:

STEP 19

Given $f(x) = x^2$ and $g(x) = \sqrt{1-x^2}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt{1-x^2}) = (\sqrt{1-x^2})^2 = 1-x^2$

STEP 20

Determine the domain of $f(g(x)) = 1-x^2$ :
The function $1-x^2$ is defined for all real numbers. However, $g(x) = \sqrt{1-x^2}$ requires $1-x^2 \geq 0$ , i.e., $x^2 \leq 1$ , which gives $-1 \leq x \leq 1$ . Therefore, the domain of $f(g(x))$ is $-1 \leq x \leq 1$ .

STEP 21

Find $g(f(x))$ :
$g(f(x)) = g(x^2) = \sqrt{1-(x^2)^2} = \sqrt{1-x^4}$

STEP 22

Determine the domain of $g(f(x)) = \sqrt{1-x^4}$ :
The expression $\sqrt{1-x^4}$ is defined when $1-x^4 \geq 0$ , i.e., $x^4 \leq 1$ , which gives $-1 \leq x \leq 1$ . Therefore, the domain of $g(f(x))$ is $-1 \leq x \leq 1$ .
Exercise 20:

STEP 23

Given $f(x) = x^3$ and $g(x) = \sqrt[3]{1-x^3}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt[3]{1-x^3}) = (\sqrt[3]{1-x^3})^3 = 1-x^3$

STEP 24

Determine the domain of $f(g(x)) = 1-x^3$ :
The function $1-x^3$ is defined for all real numbers. Therefore, the domain of $f(g(x))$ is all real numbers, $\mathbb{R}$ .

STEP 25

Find $g(f(x))$ :
$g(f(x)) = g(x^3) = \sqrt[3]{1-(x^3)^3} = \sqrt[3]{1-x^9}$

STEP 26

Determine the domain of $g(f(x)) = \sqrt[3]{1-x^9}$ :
The cube root function is defined for all real numbers. Therefore, the domain of $g(f(x))$ is all real numbers, $\mathbb{R}$ .
Exercise 21:

STEP 27

Given $f(x) = \frac{1}{2x}$ and $g(x) = \frac{1}{3x}$ , find $f(g(x))$ :
$f(g(x)) = f\left(\frac{1}{3x}\right) = \frac{1}{2 \cdot \frac{1}{3x}} = \frac{3x}{2}$

STEP 28

Determine the domain of $f(g(x)) = \frac{3x}{2}$ :
The function $\frac{3x}{2}$ is defined for all real numbers except where $x = 0$ , since $g(x) = \frac{1}{3x}$ is undefined at $x = 0$ . Therefore, the domain of $f(g(x))$ is all real numbers except $x = 0$ .

STEP 29

Find $g(f(x))$ :
$g(f(x)) = g\left(\frac{1}{2x}\right) = \frac{1}{3 \cdot \frac{1}{2x}} = \frac{2x}{3}$

STEP 30

Determine the domain of $g(f(x)) = \frac{2x}{3}$ :
The function $\frac{2x}{3}$ is defined for all real numbers except where $x = 0$ , since $f(x) = \frac{1}{2x}$ is undefined at $x = 0$ . Therefore, the domain of $g(f(x))$ is all real numbers except $x = 0$ .
Exercise 22:

STEP 31

Given $f(x) = \frac{1}{x+1}$ and $g(x) = \frac{1}{x-1}$ , find $f(g(x))$ :
$f(g(x)) = f\left(\frac{1}{x-1}\right) = \frac{1}{\frac{1}{x-1} + 1} = \frac{x-1}{1 + (x-1)} = \frac{x-1}{x}$

STEP 32

Determine the domain of $f(g(x)) = \frac{x-1}{x}$ :
The expression $\frac{x-1}{x}$ is undefined when $x = 0$ and when $x = 1$ (since $g(x) = \frac{1}{x-1}$ is undefined at $x = 1$ ). Therefore, the domain of $f(g(x))$ is all real numbers except $x = 0$ and $x = 1$ .

STEP 33

Find $g(f(x))$ :
$g(f(x)) = g\left(\frac{1}{x+1}\right) = \frac{1}{\frac{1}{x+1} - 1} = \frac{x+1}{1 - (x+1)} = \frac{x+1}{-x}$

STEP 34

Determine the domain of $g(f(x)) = \frac{x+1}{-x}$ :
The expression $\frac{x+1}{-x}$ is undefined when $x = 0$ and when $x = -1$ (since $f(x) = \frac{1}{x+1}$ is undefined at $x = -1$ ). Therefore, the domain of $g(f(x))$ is all real numbers except $x = 0$ and $x = -1$ .

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Studdy Solution

STEP 1

1. We are given pairs of functions f(x) f(x) f(x) and g(x) g(x) g(x) for each exercise.2. We need to find the compositions f(g(x)) f(g(x)) f(g(x)) and g(f(x)) g(f(x)) g(f(x)).3. We need to determine the domain of each composition function.

STEP 2

1. For each exercise, find f(g(x)) f(g(x)) f(g(x)).2. Determine the domain of f(g(x)) f(g(x)) f(g(x)).3. For each exercise, find g(f(x)) g(f(x)) g(f(x)).4. Determine the domain of g(f(x)) g(f(x)) g(f(x)).Let's go through each exercise one by one.**Exercise 15:**

STEP 3

Given f(x)=3x+2 f(x) = 3x + 2 f(x)=3x+2 and g(x)=x−1 g(x) = x - 1 g(x)=x−1, find f(g(x)) f(g(x)) f(g(x)):f(g(x))=f(x−1)=3(x−1)+2=3x−3+2=3x−1 f(g(x)) = f(x-1) = 3(x-1) + 2 = 3x - 3 + 2 = 3x - 1 f(g(x))=f(x−1)=3(x−1)+2=3x−3+2=3x−1

STEP 4

Determine the domain of f(g(x))=3x−1 f(g(x)) = 3x - 1 f(g(x))=3x−1:The function 3x−1 3x - 1 3x−1 is a linear function, so its domain is all real numbers, R \mathbb{R} R.

STEP 5

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(3x+2)=(3x+2)−1=3x+1 g(f(x)) = g(3x + 2) = (3x + 2) - 1 = 3x + 1 g(f(x))=g(3x+2)=(3x+2)−1=3x+1

STEP 6

Determine the domain of g(f(x))=3x+1 g(f(x)) = 3x + 1 g(f(x))=3x+1:The function 3x+1 3x + 1 3x+1 is a linear function, so its domain is all real numbers, R \mathbb{R} R. **Exercise 16:**

STEP 7

Given f(x)=x2−1 f(x) = x^2 - 1 f(x)=x2−1 and g(x)=1x−1 g(x) = \frac{1}{x-1} g(x)=x−11​, find f(g(x)) f(g(x)) f(g(x)):f(g(x))=f(1x−1)=(1x−1)2−1 f(g(x)) = f\left(\frac{1}{x-1}\right) = \left(\frac{1}{x-1}\right)^2 - 1 f(g(x))=f(x−11​)=(x−11​)2−1

STEP 8

Determine the domain of f(g(x))=(1x−1)2−1 f(g(x)) = \left(\frac{1}{x-1}\right)^2 - 1 f(g(x))=(x−11​)2−1:The expression 1x−1\frac{1}{x-1}x−11​ is undefined when x=1x = 1x=1. Therefore, the domain of f(g(x)) f(g(x)) f(g(x)) is all real numbers except x=1 x = 1 x=1.

STEP 9

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(x2−1)=1x2−1−1=1x2−2 g(f(x)) = g(x^2 - 1) = \frac{1}{x^2 - 1 - 1} = \frac{1}{x^2 - 2} g(f(x))=g(x2−1)=x2−1−11​=x2−21​

STEP 10

STEP 11

Given f(x)=x2−2 f(x) = x^2 - 2 f(x)=x2−2 and g(x)=x+1 g(x) = \sqrt{x+1} g(x)=x+1​, find f(g(x)) f(g(x)) f(g(x)):f(g(x))=f(x+1)=(x+1)2−2=x+1−2=x−1 f(g(x)) = f(\sqrt{x+1}) = (\sqrt{x+1})^2 - 2 = x + 1 - 2 = x - 1 f(g(x))=f(x+1​)=(x+1​)2−2=x+1−2=x−1

STEP 12

STEP 13

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(x2−2)=x2−2+1=x2−1 g(f(x)) = g(x^2 - 2) = \sqrt{x^2 - 2 + 1} = \sqrt{x^2 - 1} g(f(x))=g(x2−2)=x2−2+1​=x2−1​

STEP 14

STEP 15

Given f(x)=1x−1 f(x) = \frac{1}{x-1} f(x)=x−11​ and g(x)=x g(x) = \sqrt{x} g(x)=x​, find f(g(x)) f(g(x)) f(g(x)):f(g(x))=f(x)=1x−1 f(g(x)) = f(\sqrt{x}) = \frac{1}{\sqrt{x} - 1} f(g(x))=f(x​)=x​−11​

STEP 16

STEP 17

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(1x−1)=1x−1 g(f(x)) = g\left(\frac{1}{x-1}\right) = \sqrt{\frac{1}{x-1}} g(f(x))=g(x−11​)=x−11​​

STEP 18

STEP 19

Given f(x)=x2 f(x) = x^2 f(x)=x2 and g(x)=1−x2 g(x) = \sqrt{1-x^2} g(x)=1−x2​, find f(g(x)) f(g(x)) f(g(x)):f(g(x))=f(1−x2)=(1−x2)2=1−x2 f(g(x)) = f(\sqrt{1-x^2}) = (\sqrt{1-x^2})^2 = 1-x^2 f(g(x))=f(1−x2​)=(1−x2​)2=1−x2

STEP 20

STEP 21

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(x2)=1−(x2)2=1−x4 g(f(x)) = g(x^2) = \sqrt{1-(x^2)^2} = \sqrt{1-x^4} g(f(x))=g(x2)=1−(x2)2​=1−x4​

STEP 22

STEP 23

Given f(x)=x3 f(x) = x^3 f(x)=x3 and g(x)=1−x33 g(x) = \sqrt[3]{1-x^3} g(x)=31−x3​, find f(g(x)) f(g(x)) f(g(x)):f(g(x))=f(1−x33)=(1−x33)3=1−x3 f(g(x)) = f(\sqrt[3]{1-x^3}) = (\sqrt[3]{1-x^3})^3 = 1-x^3 f(g(x))=f(31−x3​)=(31−x3​)3=1−x3

STEP 24

Determine the domain of f(g(x))=1−x3 f(g(x)) = 1-x^3 f(g(x))=1−x3:The function 1−x3 1-x^3 1−x3 is defined for all real numbers. Therefore, the domain of f(g(x)) f(g(x)) f(g(x)) is all real numbers, R \mathbb{R} R.

STEP 25

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(x3)=1−(x3)33=1−x93 g(f(x)) = g(x^3) = \sqrt[3]{1-(x^3)^3} = \sqrt[3]{1-x^9} g(f(x))=g(x3)=31−(x3)3​=31−x9​

STEP 26

Determine the domain of g(f(x))=1−x93 g(f(x)) = \sqrt[3]{1-x^9} g(f(x))=31−x9​:The cube root function is defined for all real numbers. Therefore, the domain of g(f(x)) g(f(x)) g(f(x)) is all real numbers, R \mathbb{R} R. **Exercise 21:**

STEP 27

Given f(x)=12x f(x) = \frac{1}{2x} f(x)=2x1​ and g(x)=13x g(x) = \frac{1}{3x} g(x)=3x1​, find f(g(x)) f(g(x)) f(g(x)):f(g(x))=f(13x)=12⋅13x=3x2 f(g(x)) = f\left(\frac{1}{3x}\right) = \frac{1}{2 \cdot \frac{1}{3x}} = \frac{3x}{2} f(g(x))=f(3x1​)=2⋅3x1​1​=23x​

STEP 28

STEP 29

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(12x)=13⋅12x=2x3 g(f(x)) = g\left(\frac{1}{2x}\right) = \frac{1}{3 \cdot \frac{1}{2x}} = \frac{2x}{3} g(f(x))=g(2x1​)=3⋅2x1​1​=32x​

STEP 30

STEP 31

STEP 32

STEP 33

Find g(f(x)) g(f(x)) g(f(x)):g(f(x))=g(1x+1)=11x+1−1=x+11−(x+1)=x+1−x g(f(x)) = g\left(\frac{1}{x+1}\right) = \frac{1}{\frac{1}{x+1} - 1} = \frac{x+1}{1 - (x+1)} = \frac{x+1}{-x} g(f(x))=g(x+11​)=x+11​−11​=1−(x+1)x+1​=−xx+1​

STEP 34

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1. We are given pairs of functions $f(x)$ and $g(x)$ for each exercise.
2. We need to find the compositions $f(g(x))$ and $g(f(x))$ .
3. We need to determine the domain of each composition function.

1. For each exercise, find $f(g(x))$ .
2. Determine the domain of $f(g(x))$ .
3. For each exercise, find $g(f(x))$ .
4. Determine the domain of $g(f(x))$ .
Let's go through each exercise one by one.
Exercise 15:

Given $f(x) = 3x + 2$ and $g(x) = x - 1$ , find $f(g(x))$ :
$f(g(x)) = f(x-1) = 3(x-1) + 2 = 3x - 3 + 2 = 3x - 1$

Determine the domain of $f(g(x)) = 3x - 1$ :
The function $3x - 1$ is a linear function, so its domain is all real numbers, $\mathbb{R}$ .

Find $g(f(x))$ :
$g(f(x)) = g(3x + 2) = (3x + 2) - 1 = 3x + 1$

Determine the domain of $g(f(x)) = 3x + 1$ :
The function $3x + 1$ is a linear function, so its domain is all real numbers, $\mathbb{R}$ .
Exercise 16:

Given $f(x) = x^2 - 1$ and $g(x) = \frac{1}{x-1}$ , find $f(g(x))$ :
$f(g(x)) = f\left(\frac{1}{x-1}\right) = \left(\frac{1}{x-1}\right)^2 - 1$

Determine the domain of $f(g(x)) = \left(\frac{1}{x-1}\right)^2 - 1$ :
The expression $\frac{1}{x-1}$ is undefined when $x = 1$ . Therefore, the domain of $f(g(x))$ is all real numbers except $x = 1$ .

Find $g(f(x))$ :
$g(f(x)) = g(x^2 - 1) = \frac{1}{x^2 - 1 - 1} = \frac{1}{x^2 - 2}$

Given $f(x) = x^2 - 2$ and $g(x) = \sqrt{x+1}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt{x+1}) = (\sqrt{x+1})^2 - 2 = x + 1 - 2 = x - 1$

Find $g(f(x))$ :
$g(f(x)) = g(x^2 - 2) = \sqrt{x^2 - 2 + 1} = \sqrt{x^2 - 1}$

Given $f(x) = \frac{1}{x-1}$ and $g(x) = \sqrt{x}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt{x}) = \frac{1}{\sqrt{x} - 1}$

Find $g(f(x))$ :
$g(f(x)) = g\left(\frac{1}{x-1}\right) = \sqrt{\frac{1}{x-1}}$

Given $f(x) = x^2$ and $g(x) = \sqrt{1-x^2}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt{1-x^2}) = (\sqrt{1-x^2})^2 = 1-x^2$

Find $g(f(x))$ :
$g(f(x)) = g(x^2) = \sqrt{1-(x^2)^2} = \sqrt{1-x^4}$

Given $f(x) = x^3$ and $g(x) = \sqrt[3]{1-x^3}$ , find $f(g(x))$ :
$f(g(x)) = f(\sqrt[3]{1-x^3}) = (\sqrt[3]{1-x^3})^3 = 1-x^3$

Determine the domain of $f(g(x)) = 1-x^3$ :
The function $1-x^3$ is defined for all real numbers. Therefore, the domain of $f(g(x))$ is all real numbers, $\mathbb{R}$ .

Find $g(f(x))$ :
$g(f(x)) = g(x^3) = \sqrt[3]{1-(x^3)^3} = \sqrt[3]{1-x^9}$

Determine the domain of $g(f(x)) = \sqrt[3]{1-x^9}$ :
The cube root function is defined for all real numbers. Therefore, the domain of $g(f(x))$ is all real numbers, $\mathbb{R}$ .
Exercise 21:

Given $f(x) = \frac{1}{2x}$ and $g(x) = \frac{1}{3x}$ , find $f(g(x))$ :
$f(g(x)) = f\left(\frac{1}{3x}\right) = \frac{1}{2 \cdot \frac{1}{3x}} = \frac{3x}{2}$

Find $g(f(x))$ :
$g(f(x)) = g\left(\frac{1}{2x}\right) = \frac{1}{3 \cdot \frac{1}{2x}} = \frac{2x}{3}$

Find $g(f(x))$ :
$g(f(x)) = g\left(\frac{1}{x+1}\right) = \frac{1}{\frac{1}{x+1} - 1} = \frac{x+1}{1 - (x+1)} = \frac{x+1}{-x}$