Find \((g \circ f)(x)\)
Substitute \(f(x)\) into \(g(x)\)
To find \((g \circ f)(x)\), substitute \(f(x) = \frac{3}{1-7x}\) into \(g(x) = \frac{1}{x}\). This gives us:
\[
g(f(x)) = g\left(\frac{3}{1-7x}\right) = \frac{1}{\frac{3}{1-7x}}
\]
Simplify the expression
Simplify \(\frac{1}{\frac{3}{1-7x}}\) to get:
\[
\frac{1-7x}{3}
\]
\(\boxed{(g \circ f)(x) = \frac{1-7x}{3}}\)
Find the domain of \((f \circ g)(x)\)
Identify where the denominator is zero
The function \((f \circ g)(x) = \frac{3x}{x-7}\) has a denominator of \(x-7\). Set the denominator not equal to zero:
\[
x-7 \neq 0 \implies x \neq 7
\]
State the domain
The domain of \((f \circ g)(x)\) is all real numbers except 7:
\[
(-\infty, 7) \cup (7, \infty)
\]
\(\boxed{\text{Domain of } (f \circ g)(x): (-\infty, 7) \cup (7, \infty)}\)
Find the domain of \((g \circ f)(x)\)
Consider the domain of the inner function \(f(x)\)
The function \(f(x) = \frac{3}{1-7x}\) has a denominator of \(1-7x\). Set the denominator not equal to zero:
\[
1-7x \neq 0 \implies x \neq \frac{1}{7}
\]
State the domain
The domain of \((g \circ f)(x)\) is all real numbers except \(\frac{1}{7}\):
\[
(-\infty, \frac{1}{7}) \cup (\frac{1}{7}, \infty)
\]
\(\boxed{\text{Domain of } (g \circ f)(x): (-\infty, \frac{1}{7}) \cup (\frac{1}{7}, \infty)}\)
\((g \circ f)(x) = \frac{1-7x}{3}\)
Domain of \((f \circ g)(x): (-\infty, 7) \cup (7, \infty)\)
Domain of \((g \circ f)(x): (-\infty, \frac{1}{7}) \cup (\frac{1}{7}, \infty)\)
Answered
