Math Snap

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

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 $x = -1$.