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
As we are given that $y=e^x$
Let us consider $f(x)=y=e^x$
Write formula 11.
Now, $\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$
$f'(x)=e^x$ and $f''(x)=e^x$
Thus, $\kappa(x)=\dfrac{|e^x|}{[1+(e^x)^2]^{3/2}}=\dfrac{e^x}{(1+e^{2x})^{3/2}}$
Since, $\kappa'(x)=\dfrac{e^x}{(1+e^{2x})^{5/2}}(1-2e^{2x})$
Hence, $\kappa (x)$ has maximum at $(-\frac{1}{2} \ln 2,\frac{1}{\sqrt 2})$ and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$
