Is there a way of calculating where the apex of an exponential lies? There's probably a deeper / more mathematical way of explaining what this is exactly. The image hopefully demonstrates what I mean. I'm not sure if derivatives help and perhaps numerical methods need to be used?
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If we look for the maximum of the curvature of the curve $y=e^{ax}$, we have $$\kappa=\frac{y''}{(1+(y')^2)^{3/2}}=\frac{a^2 e^{a x}}{\left(1+a^2 e^{2 a x}\right)^{3/2}}$$ $$\frac{d\kappa}{dx}=\frac{a^3e^{ax}(1-2a^2e^{2ax})}{\left(1+a^2 e^{2 a x}\right)^{5/2}}$$ So, the maximum curvature is obtained for $$x=-\frac{\log(2a^2)}{2a}$$ for which $$\kappa=\frac{2 a}{3 \sqrt{3}}$$ At this point, the second derivative $$\frac{d^2\kappa}{dx^2}=\frac{a^4 e^{a x} \left(4 a^4 e^{4 a x}-10 a^2 e^{2 a x}+1\right)}{\left(1+a^2 e^{2 a x}\right)^{7/2}}=-\frac{8 a^3}{9 \sqrt{3}}$$ confirms the fact that it is a maximum.
For $a=1$, the maximum curvature would happen at $x=-\frac{\log (2)}{2}$.
But, as you say, may be the maximum curvature is not what you are looking for.
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I guess exponential is too crazy for that kind of stuff - it's growing - well, exponentially - so basically, always growing more than ever before - not just getting bigger - but getting more bigger than it ever got bigger before. All I'm saying is that the derivative always grows, i.e. that the second derivative is positive. Problem is, all derivatives are positive anyway ... So it's also getting more more bigger than it ever got more bigger before, and so on ...
At the end of the day, such apexes as searched are everywhere, given that the slope is always steeper after any point than it was before, by a margin that was never so great before, ...
You can compute the maximum of the curvature of the curve $y=e^x$. This can be done by the formula $$\kappa=\frac{y''}{(1+(y')^2)^{3/2}}=\frac{e^x}{(1+e^{2x})^{3/2}}.$$ A simple computation shows that this is maximal for $x=-\frac{\ln 2}{2}$ (after edit).
However, for the function $A\,e^{Bx}$ where $A,B>0$, the result may be different and the computed quantity is not invariant with respect to the "scaling" of your graph. Note that you have a different scale in the $x$- and $y$-axis in your picture.
By a "measurement on the screen" it seems to me that your 4 units on the $x$-axis correspond to $100$ on the $y$-axis in your graph. So you really draw the function $\frac{1}{25} e^x$ (using an equi-scaled graph) which leads to the maximal curvature in $\frac{-\ln 2}{2}+\ln 25\simeq 2.87$ which is already approximately 3 as you want.