At what point is the curvature of $y=-\ln(x)$ maximal? I used the equation 
$$k = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}.$$
Just by looking at its would think the max is at $x = 1, y = 0$.
What is the correct answer?
 A: $f(x) = -\ln{x} \implies f'(x) = -1/x \implies f''(x) = 1/x^2$.
Then
$$k(x)=x (1+x^2)^{-3/2}$$
The derivative of $k$ is
$$k'(x) = \frac{1-2 x^2}{\left(x^2+1\right)^{5/2}}$$
which is zero when $x=1/\sqrt{2}$.  You can show that $k''(1/\sqrt{2}) < 0$, so the curvature is maximized at $x=1/\sqrt{2}$.
A: Alternately, you can maximize the curvature of the inverse function $$g(x)=e^{-x},$$ then simply swap the $x$ and $y$ coordinates to see where $f(x)$ has maximum curvature.
The curvature of $g$ will be $$\begin{align}\frac{|g''(x)|}{\left(1+\left(g'(x)\right)^2\right)^{\frac32}} &= \frac{\left|e^{-x}\right|}{\left(1+\left(-e^{-x}\right)^2\right)^{\frac32}}\\ &= \frac{e^{-x}}{\left(1+e^{-2x}\right)^{\frac32}}\\ &= \frac{e^{-x}}{e^{-3x}\left(e^{2x}+1\right)^{\frac32}}\\ &= \frac{e^{2x}}{\left(e^{2x}+1\right)^{\frac32}}.\end{align}$$
The derivative of the curvature is $$\frac{2e^{2x}-e^{4x}}{\left(e^{2x}+1\right)^{\frac52}},$$ which is zero if and only if $e^{2x}=2$ if and only if $2x=\ln(2)$ if and only if $x=\frac12\ln(2)$. Since $g\left(\frac12\ln(2)\right)=\frac1{\sqrt{2}},$ then $y=g(x)$ has maximum curvature when $x=\frac12\ln(2)$ and $y=\frac1{\sqrt{2}}.$ Thus, $y=-\ln(x)$ has maximum curvature when $x=\frac1{\sqrt{2}}$ and $y=\frac12\ln(2).$
It's a bit out of our way, in this case, and not every function has an inverse, but sometimes a given function will have an inverse that's easier to deal with, so this is a trick that's worth keeping in our toolbox.
