Answer
The value(s) of $\alpha$ such that the curvature of $y = {{\rm{e}}^{\alpha x}}$ at $x=0$ is as large as possible are $\alpha = \pm \sqrt 2 $.
Work Step by Step
We have $y = {{\rm{e}}^{\alpha x}}$.
Write $y = f\left( x \right) = {{\rm{e}}^{\alpha x}}$. So, $f'\left( x \right) = \alpha {{\rm{e}}^{\alpha x}}$ and $f{\rm{''}}\left( x \right) = {\alpha ^2}{{\rm{e}}^{\alpha x}}$.
By Eq. (5) of Theorem 2, the curvature is given by
$\kappa \left( x \right) = \frac{{\left| {f{\rm{''}}\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( x \right) = \frac{{\left| {{\alpha ^2}{{\rm{e}}^{\alpha x}}} \right|}}{{{{\left( {1 + {\alpha ^2}{{\rm{e}}^{2\alpha x}}} \right)}^{3/2}}}}$
At $x=0$, we have $\kappa \left( 0 \right) = \frac{{{\alpha ^2}}}{{{{\left( {1 + {\alpha ^2}} \right)}^{3/2}}}}$
Next, we find the values of $\alpha$ that make $\kappa \left( 0 \right)$ maximum.
We write $g\left( \alpha \right) = \kappa \left( 0 \right) = \frac{{{\alpha ^2}}}{{{{\left( {1 + {\alpha ^2}} \right)}^{3/2}}}}$.
The derivatives are
$g'\left( \alpha \right) = \frac{{\alpha \left( {2 - {\alpha ^2}} \right)}}{{{{\left( {1 + {\alpha ^2}} \right)}^{5/2}}}}$, ${\ \ }$ $g{\rm{''}}\left( \alpha \right) = \frac{{2{\alpha ^4} - 11{\alpha ^2} + 2}}{{{{\left( {1 + {\alpha ^2}} \right)}^{7/2}}}}$
We solve the equation $g'\left( \alpha \right) = 0$ to find the critical points. So,
$\frac{{\alpha \left( {2 - {\alpha ^2}} \right)}}{{{{\left( {1 + {\alpha ^2}} \right)}^{5/2}}}} = 0$
The denominator is nonvanishing, so
$\alpha \left( {2 - {\alpha ^2}} \right) = 0$
The solutions are $\alpha=0$ and $\alpha = \pm \sqrt 2 $.
Substituting $\alpha=0$ in $g{\rm{''}}\left( \alpha \right)$ gives $g{\rm{''}}\left( 0 \right) = 2$. Since $g{\rm{''}}\left( 0 \right) > 0$, the curvature is minimum at $\alpha=0$.
Substituting $\alpha = \sqrt 2 $ in $g{\rm{''}}\left( \alpha \right)$ gives $g{\rm{''}}\left( {\sqrt 2 } \right) = - \frac{4}{{9\sqrt 3 }}$. Since $g{\rm{''}}\left( 0 \right) < 0$, the curvature is maximum at $\alpha = \sqrt 2 $.
Substituting $\alpha = - \sqrt 2 $ in $g{\rm{''}}\left( \alpha \right)$ gives $g{\rm{''}}\left( { - \sqrt 2 } \right) = - \frac{4}{{9\sqrt 3 }}$. Since $g{\rm{''}}\left( 0 \right) < 0$, the curvature is maximum at $\alpha = - \sqrt 2 $.
Hence, the value(s) of $\alpha$ such that the curvature of $y = {{\rm{e}}^{\alpha x}}$ at $x=0$ is as large as possible are $\alpha = \pm \sqrt 2 $.
