Mathematics

The key insight is that the derivative of an inverse function is given by $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$. Since $f(2) = 5$, we know $f^{-1}(5) = 2$. But $f'(2) = 0$, which makes the denominator zero—so the formula breaks down and the inverse has a vertical tangent at $x = 5$. A vertical tangent means the inverse function isn't differentiable there.

The key insight is that the derivative of an inverse function is given by $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$. Since $f(2) = 5$, we know $f^{-1}(5) = 2$. But $f'(2) = 0$, which makes the denominator zero—so the formula breaks down and the inverse has a vertical tangent at $x = 5$. A vertical tangent means the inverse function isn't differentiable there.