How to find the maximum curvature of $y=e^x$?

Question

So I found the curvature to be $K = \dfrac{e^{x}}{(1+e^{2x})^{3/2}}$ but I don't know how to maximize this?

Answer 1

$$K(x)=\frac{e^x}{(1+e^{2x})^{3/2}} = \frac{1}{\left(e^{-2x/3}+e^{4x/3}\right)^{3/2}}$$ hence to maximize $K(x)$ is the same as minimizing: $$ f(x) = e^{-2x/3}+e^{4x/3}. $$ By solving $f'(x)=0$ we get that $x=-\frac{\log 2}{2}$ is the only stationary point of $f(x)$, hence the only stationary point of $K(x)$, and: $$ K(x) \leq \frac{2}{3\sqrt{3}}.$$

Answer 2

We can differentiate and set equal to zero to get

$$\frac{e^x-2e^{3x}}{(1+e^{2x})^{5/2}}=0\implies e^x=2e^{3x}\implies\frac{1}{2}=e^{2x}.\tag{1}$$

Then solve to obtain $x=-\frac{1}{2}\log 2$.

Note - the various algebraic operations in (1) are possible since $e^x\neq 0$ for all $x\in\mathbb{R}$. For future similar questions, it may be necessary to determine maxima/minima using the second derivative. So called "saddle" points may also need addressing.