1. The problem asks to find $f^{-1}(5)$ where $f(x) = x^2 + 1$. 2. The function $f(x) = x^2 + 1$ maps $x$ to $x^2 + 1$. The inverse function $f^{-1}(y)$ gives the value of $x$ such that $f(x) = y$. 3. To find $f^{-1}(5)$, solve the equation $x^2 + 1 = 5$ for $x$. 4. Subtract 1 from both sides: $$x^2 + 1 = 5 \\ x^2 = 5 - 1 \\ x^2 = 4$$ 5. Take the square root of both sides: $$x = \pm \sqrt{4}$$ 6. Simplify the square root: $$x = \pm 2$$ 7. Since $f(x) = x^2 + 1$ is not one-to-one over all real numbers, the inverse is not a function unless we restrict the domain. Usually, the principal (non-negative) root is taken for the inverse function. 8. Therefore, $f^{-1}(5) = 2$. Final answer: $f^{-1}(5) = 2$