University Calculus: Early Transcendentals (3rd Edition)

$f\circ g\circ h\ =\ \frac{8-3x}{7-2x}$
$f(x)=\frac{x+2}{3-x}$ $g(x)=\frac{x^2}{x^2+1}$ $h(x)=\sqrt{2-x}$ $f\circ g\circ h=f(g(h(x)))$ 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: $=\frac{8-3x}{7-2x}$