Answer
$f\circ g\circ h\ =\ \frac{8-3x}{7-2x}$
Work Step by Step
$f(x)=\frac{x+2}{3-x}$
$g(x)=\frac{x^2}{x^2+1}$
$h(x)=\sqrt{2-x}$
$f\circ g\circ h=f(g(h(x)))$
First evalulate the composition of the inner two function as:
$g\circ h=g(h(x))$
$=\frac{(\sqrt{2-x})^2}{(\sqrt{2-x})^2+1}$
Simplify:
$=\frac{2-x}{3-x}$
Now find the remaining part as:
$f\circ g\circ h=f(g(h(x)))$
$f(\frac{2-x}{3-x})=\frac{\frac{2-x}{3-x}+2}{3-\frac{2-x}{3-x}}$
Simplify:
$=\frac{8-3x}{7-2x}$
