Answer
The maximum of the curvature occurs at $x = - \frac{1}{2}\ln 2$.
Thus, the point of maximum curvature is $\left( {x,\kappa \left( x \right)} \right) = \left( { - \frac{1}{2}\ln 2,\frac{2}{{3\sqrt 3 }}} \right)$ on the $x\kappa$-plane.
Work Step by Step
We have $y = {{\rm{e}}^x}$.
Write $y = f\left( x \right) = {{\rm{e}}^x}$. So, $f'\left( x \right) = {{\rm{e}}^x}$ and $f{\rm{''}}\left( x \right) = {{\rm{e}}^x}$.
By Eq. (5) of Theorem 2, the curvature is given by
$\kappa \left( x \right) = \frac{{\left| {f{\rm{''}}\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( x \right) = \frac{{\left| {{{\rm{e}}^x}} \right|}}{{{{\left( {1 + {{\rm{e}}^{2x}}} \right)}^{3/2}}}} = \frac{{{{\rm{e}}^x}}}{{{{\left( {1 + {{\rm{e}}^{2x}}} \right)}^{3/2}}}}$
The derivatives are
$\kappa '\left( x \right) = \frac{{ - 2{{\rm{e}}^{3x}} + {{\rm{e}}^x}}}{{{{\left( {1 + {{\rm{e}}^{2x}}} \right)}^{5/2}}}}$, ${\ \ }$ $\kappa {\rm{''}}\left( x \right) = \frac{{4{{\rm{e}}^{5x}} - 10{{\rm{e}}^{3x}} + {{\rm{e}}^x}}}{{{{\left( {1 + {{\rm{e}}^{2x}}} \right)}^{7/2}}}}$
We solve the equation $\kappa '\left( x \right) = 0$ to find the critical points. So,
$\frac{{ - 2{{\rm{e}}^{3x}} + {{\rm{e}}^x}}}{{{{\left( {1 + {{\rm{e}}^{2x}}} \right)}^{5/2}}}} = 0$
The denominator is nonvanishing, so
$ - 2{{\rm{e}}^{3x}} + {{\rm{e}}^x} = 0$
${{\rm{e}}^x}\left( {1 - 2{{\rm{e}}^{2x}}} \right) = 0$
Since ${{\rm{e}}^x} \ne 0$, so $1 - 2{{\rm{e}}^{2x}} = 0$.
${{\rm{e}}^{2x}} = \frac{1}{2}$, ${\ \ \ }$ $2x = \ln \frac{1}{2}$
Thus, the critical point is at $x = - \frac{1}{2}\ln 2$.
Substituting $x = - \frac{1}{2}\ln 2$ in $\kappa {\rm{''}}\left( x \right)$ gives $\kappa {\rm{''}}\left( { - \frac{1}{2}\ln 2} \right) = - \frac{8}{{9\sqrt 3 }}$. Since $\kappa {\rm{''}}\left( { - \frac{1}{2}\ln 2} \right) < 0$, thus, the maximum of the curvature occurs at $x = - \frac{1}{2}\ln 2$.
Substituting $x = - \frac{1}{2}\ln 2$ in $\kappa \left( x \right)$ gives $\kappa \left( { - \frac{1}{2}\ln 2} \right) = \frac{2}{{3\sqrt 3 }}$. Thus, the point of maximum curvature is $\left( {x,\kappa \left( x \right)} \right) = \left( { - \frac{1}{2}\ln 2,\frac{2}{{3\sqrt 3 }}} \right)$ on the $x\kappa$-plane.
