Question

PLEASE HELP ME!!!!!!!!!!!!! Let f(x)=x^2+7 and g(x)=x-3/x find (f o g) (-8)

Answer 1

Answer:

$f{\circ }g(-8)=\frac{781}{8}$

Step-by-step explanation:

We have been given two function $f(x)=x^2+7$ and $g(x)=\frac{x-3}{x}$. We are asked to find $f{\circ }g(-8)$.

By the definition of composite functions $f{\circ}g(x)=f(x)\cdot g(x)$.

$f{\circ }g(-8)=f(-8)\cdot g(-8)$

$f(-8)=(-8)^2+7$

$f(-8)=64+7$

$f(-8)=71$

$g(-8)=\frac{-8-3}{-8}$

$g(-8)=\frac{-11}{-8}$

$g(-8)=\frac{11}{8}$

$f{\circ }g(-8)=71\cdot \frac{11}{8}$

$f{\circ }g(-8)=\frac{781}{8}$

Therefore, $f{\circ}g(x)=\frac{781}{8}$.

Answer 2

X=-56 that is the answer