Question

for the real-valued functions $g(x)=\frac{x-4}{x+1}$ and $h(x)=4x-13$, find the composition $g \circ h$. also, specify its domain using interval notation.
$(g \circ h)(x) = \square$
domain of $g \circ h$: $\square$

Explanation:

Step1: Define composition $g\circ h$

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

Step2: Substitute $h(x)$ into $g$

$g(4x-13) = \frac{(4x-13)-4}{(4x-13)+1}$

Step3: Simplify numerator and denominator

$\frac{4x-13-4}{4x-13+1} = \frac{4x-17}{4x-12}$

Step4: Simplify the fraction

$\frac{4x-17}{4(x-3)}$

Step5: Find domain restrictions

First, denominator of $g\circ h$ cannot be 0:
$4x-12
eq 0 \implies x
eq 3$
Also, $h(x)$ must be in domain of $g$: domain of $g$ excludes $x=-1$, so $4x-13
eq -1$
$4x
eq 12 \implies x
eq 3$ (same restriction)

Step6: Write domain in interval notation

$(-\infty, 3) \cup (3, \infty)$

Answer:

$(g \cdot h)(x) = \boldsymbol{\frac{4x-17}{4x-12}}$
Domain of $g \cdot h$: $\boldsymbol{(-\infty, 3) \cup (3, \infty)}$