# Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

#### If $f\left(x\right)=\frac{4}{x+3}\:and\:g\left(x\right)=x^2-3,\:find\:f\left[g\left(x\right)\right]\:and\:g\left[f\left(x\right)\right]$

SOLUTION:

$f\left[g\left(x\right)\right]=\frac{4}{\left(x^2+3\right)+3}$

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

$g\left[f\left(x\right)\right]=\left(\frac{4}{x+3}\right)^2-3$

$=\frac{16}{\left(x+3\right)^2}-3$

$=\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}$

$=\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}$

$=\frac{16-3x^2-18x-27}{\left(x+3\right)^2}$

$=\frac{-3x^2-18x-11}{\left(x+3\right)^2}$