Question

The function $f(x)$ satisfies

\[f(\sqrt{x + 1}) = \frac{1}{x}\]for all $x \ge -1,$ $x\\eq 0.$

What is the value of $f(2)$?

Answer 1

We can start by making a substitution $y = \sqrt{x + 1}$, so that $x = y^2 - 1$. Then we have
[f(y) = f(\sqrt{y^2 - 1 + 1}) = f(y)]for all $y \ge 0$.

In particular, we can substitute $y = \sqrt{2}$ to get
[f(\sqrt{2}) = \frac{1}{(\sqrt{2})^2 - 1} = \frac{1}{1} = \boxed{1}.]

Answer 2

Setting $x= y^2-1$ gives

[f(y) = \frac{1}{y^2-1}.]

Thus, we have $f(\sqrt{2+1}) = f(\sqrt{3}) = \frac{1}{2},$ so $f(2) = \boxed{\frac{1}{3}}.$