The function \[f(\sqrt{x + 1}) = \frac{1}{x}\] satisfies
for all $x \ge -1,$ Find $f(2)$.
+171 we are given $f(\sqrt{x + 1}) = \frac{1}{x}$, but we know that $\sqrt{x + 1} \equiv 2$ which gives you $x+1=4 \ \ \ \therefore \ \ \ x=3 $
knowing that we can put that x on the other hand and for the first side we can either plug in $3$ for x, or just use $f(2)$ for the whole thing (which ill be using as its... easier.):
so you get:
$\boxed{ f(2)=\frac{1}{3}} $
