Question

Use the given functions to find f (g(x)), and give the restrictions on x.

Answer 1

Answer:

$f(g(x)) = \frac{10x-9}{3 + 3x}$

Step-by-step explanation:

Given

$f(x) = \frac{1}{x} - 3$

$g(x) = \frac{3}{x} + 3$

Required

Find f(g(x))

If: $f(x) = \frac{1}{x} - 3$

Then:

$f(g(x)) = \frac{1}{3/x + 3} -3$

Solve the denominator (take LCM)

$f(g(x)) = \frac{1}{\frac{3+3x}{x}} -3$

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

It can be solved further as:

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

$f(g(x)) = \frac{x-9 + 9x}{3 + 3x}$

$f(g(x)) = \frac{10x-9}{3 + 3x}$