Question

What is the domain of (fog)(x) if
f(x)=1/x+2 and
g(x)=4/x​

Answer 1

Assuming that

$1/x + 2 = \frac{1}{x + 2} \: not \: \frac{1}{x} + 2$

Given,

$f(x) = \frac{1}{x + 2}$

$g(x) = \frac{4}{x}$

Then

$(f \circ g)(x) = f(g(x))$

$= f( \frac{4}{x} )$

$= \frac{1}{ \frac{4}{x} + 2 }$

$= \frac{1}{ \frac{4 + 2x}{x} }$

$= \frac{x}{4 + 2x}$

Here,

$(f \circ g)(x) = \frac{x}{2x + 4}$

is only defined when

$2x + 4\neq 0$

$\implies 2x\neq - 4$

$\implies x\neq - 2$

Therefore domain of (f o g)(x),

$D_{f\circ g} = \mathbb{R} - \{ - 2\}$

$= ( - \infty, - 2)\cup( - 2, \infty)$