## Thomas' Calculus 13th Edition

$(f\circ g\circ h)(x)=\dfrac{8-3x}{7-2x}.$
We are given $f(x)=\dfrac{x+2}{3-x}$, $g(x)=\dfrac{x^2}{x^2+1}$, and $h(x)=\sqrt{2-x}$. So $(f\circ g\circ h)(x)=f(g(h(x)))=f\left(g\left(\sqrt{2-x}\right)\right)=f\left(\dfrac{\left(\sqrt{2-x}\right)^2}{\left(\sqrt{2-x}\right)^2 +1}\right)$. Simplifying, we get $f\left(\dfrac{2-x}{2-x +1}\right)=f\left(\dfrac{2-x}{3-x}\right).$ Then $f\left(\dfrac{2-x}{3-x}\right)=\dfrac{\left(\dfrac{2-x}{3-x}\right)+2}{3-\left(\dfrac{2-x}{3-x}\right)}.$ Now, we multiply the numerator and denominator of our fraction by $3-x$ to get $\dfrac{2-x+2(3-x)}{3(3-x)-(2-x)}=\dfrac{2-x+6-2x}{9-3x-2+x}=\dfrac{8-3x}{7-2x}.$