## Calculus with Applications (10th Edition)

$f[g(x)]=\displaystyle \frac{8\sqrt{3-x}}{3-x}$ $g[f(x)]=\displaystyle \frac{\sqrt{3x^{2}-8x}}{x}$
In the expression for f(x), replace x with g(x) $f[g(x)]=\displaystyle \frac{8}{g(x)}$ $=\displaystyle \frac{8}{\sqrt{3-x}}$ ... rationalize the denominator $=\displaystyle \frac{8}{\sqrt{3-x}}\cdot\frac{\sqrt{3-x}}{\sqrt{3-x}}$ $=\displaystyle \frac{8\sqrt{3-x}}{3-x}$ In the expression for g(x), replace x with f(x) $g[f(x)]=\sqrt{3-f(x)}$ $=\sqrt{3-\dfrac{8}{x}}$ $=\sqrt{\dfrac{3x-8}{x}}$ ... rationalize the denominator $=\displaystyle \frac{\sqrt{3x-8}}{\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}}$ $=\displaystyle \frac{\sqrt{3x^{2}-8x}}{x}$