Answer
(a) $-\frac{1}{8}$
(b) $-\frac{1}{28}$
(c) $\frac{\sqrt x}{x^2-18x}~~$, with $x\geq0$
Work Step by Step
$f(x)=\frac{\sqrt {x+9}}{x^2-81}$
(a) $f(7)=\frac{\sqrt {7+9}}{7^2-81}=\frac{4}{-32}=-\frac{1}{8}$
(b) $f(-5)=\frac{\sqrt {-5+9}}{(-5)^2-81}=\frac{2}{-56}=-\frac{1}{28}$
(c) $f(x-9)=\frac{\sqrt {x-9+9}}{(x-9)^2-81}=\frac{\sqrt x}{x^2-18x+81-81}=\frac{\sqrt x}{x^2-18x}~~$, with $x\geq0$
