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For the given functions, find $(f \circ g)(x)$ and $(g \circ f)(x)$ and the domain of each. $\begin{array}{l} f(x)=\frac{8}{1-3 x}, g(x)=\frac{1}{x} \\ (f \circ g)(x)=\frac{8 x}{x-3} \end{array}$ (Simplify your answer. Use integers or fractions for any numbers in the expression.) $(g \circ f)(x)=$ $\square$ (Simplify your answer. Use integers or fractions for any numbers in the expression.)

Studdy Solution

STEP 1

1. We are given two functions $f(x) = \frac{8}{1-3x}$ and $g(x) = \frac{1}{x}$ .
2. We need to find the compositions $(f \circ g)(x)$ and $(g \circ f)(x)$ .
3. The domain of a composition is the set of all $x$ values for which the composition is defined.

STEP 2

1. Calculate $(f \circ g)(x)$ .
2. Determine the domain of $(f \circ g)(x)$ .
3. Calculate $(g \circ f)(x)$ .
4. Determine the domain of $(g \circ f)(x)$ .

STEP 3

To find $(f \circ g)(x)$ , substitute $g(x)$ into $f(x)$ :
$(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right)$
Substitute $\frac{1}{x}$ into $f(x)$ :
$f\left(\frac{1}{x}\right) = \frac{8}{1 - 3\left(\frac{1}{x}\right)}$
Simplify the expression:
$= \frac{8}{1 - \frac{3}{x}}$
Combine terms in the denominator:
$= \frac{8}{\frac{x - 3}{x}}$
Simplify by multiplying by the reciprocal:
$= 8 \cdot \frac{x}{x-3} = \frac{8x}{x-3}$

STEP 4

Determine the domain of $(f \circ g)(x) = \frac{8x}{x-3}$ :
1. $g(x) = \frac{1}{x}$ is defined for $x \neq 0$ .
2. $f(g(x)) = \frac{8x}{x-3}$ is defined for $x \neq 3$ .
The domain of $(f \circ g)(x)$ is $x \neq 0$ and $x \neq 3$ .

STEP 5

To find $(g \circ f)(x)$ , substitute $f(x)$ into $g(x)$ :
$(g \circ f)(x) = g(f(x)) = g\left(\frac{8}{1-3x}\right)$
Substitute $\frac{8}{1-3x}$ into $g(x)$ :
$g\left(\frac{8}{1-3x}\right) = \frac{1}{\frac{8}{1-3x}}$
Simplify by multiplying by the reciprocal:
$= \frac{1-3x}{8}$

STEP 6

Determine the domain of $(g \circ f)(x) = \frac{1-3x}{8}$ :
1. $f(x) = \frac{8}{1-3x}$ is defined for $1-3x \neq 0$ , which implies $x \neq \frac{1}{3}$ .
2. $g(f(x)) = \frac{1-3x}{8}$ is defined for all real numbers since there are no additional restrictions.
The domain of $(g \circ f)(x)$ is $x \neq \frac{1}{3}$ .
The compositions are: - $(f \circ g)(x) = \frac{8x}{x-3}$ with domain $x \neq 0$ and $x \neq 3$ . - $(g \circ f)(x) = \frac{1-3x}{8}$ with domain $x \neq \frac{1}{3}$ .

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