Answer
$\kappa (x)$ has maximum at $(-\frac{1}{2} \ln 2,\frac{1}{\sqrt 2})$
and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$
Work Step by Step
Given: $y=e^x$
Consider $f(x)=y=e^x$
In order to find the curvature we will have to use formula 11, such that
$\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$
$y'=e^x$
and $y''=e^x$
$\kappa(x)=\dfrac{|e^x|}{[1+(e^x)^2]^{3/2}}=\dfrac{e^x}{(1+e^{2x})^{3/2}}$
$\kappa'(x)=\dfrac{e^x}{(1+e^{2x})^{5/2}}(1-2e^{2x})$
$\kappa (x)$ has maximum at $(-\frac{1}{2} \ln 2,\frac{1}{\sqrt 2})$
and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$
