if $g(x)=\frac{x + 1}{x - 2}$ and $h(x)=4 - x$, what is the value of $(gcirc h)(-3)$?\n$\frac{8}{5}$\n$\frac{…
if $g(x)=\frac{x + 1}{x - 2}$ and $h(x)=4 - x$, what is the value of $(gcirc h)(-3)$?\n$\frac{8}{5}$\n$\frac{5}{2}$\n$\frac{15}{2}$\n$\frac{18}{5}$
Answer
Explanation:
Step1: Find the value of $h(-3)$
Substitute $x = - 3$ into $h(x)=4 - x$. $h(-3)=4-(-3)=4 + 3=7$
Step2: Find the value of $g(h(-3))$
Since $h(-3)=7$, substitute $x = 7$ into $g(x)=\frac{x + 1}{x-2}$. $g(7)=\frac{7 + 1}{7-2}=\frac{8}{5}$
Answer:
$\frac{8}{5}$