Answer
The minimum speed occurs at $t = \frac{7}{2}$:
$||{\bf{r}}'\left( {\frac{7}{2}} \right)|| = \sqrt {1 + 2{\rm{e}}} $
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,{{\rm{e}}^{t - 3}},{{\rm{e}}^{4 - t}}} \right)$.
So, ${\bf{r}}'\left( t \right) = \left( {1,{{\rm{e}}^{t - 3}}, - {{\rm{e}}^{4 - t}}} \right)$
The speed is
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,{{\rm{e}}^{t - 3}}, - {{\rm{e}}^{4 - t}}} \right)\cdot\left( {1,{{\rm{e}}^{t - 3}}, - {{\rm{e}}^{4 - t}}} \right)} $
$ = \sqrt {1 + {{\rm{e}}^{2\left( {t - 3} \right)}} + {{\rm{e}}^{2\left( {4 - t} \right)}}} $
Write $f\left( t \right) = ||{\bf{r}}'\left( t \right)|| = \sqrt {1 + {{\rm{e}}^{2\left( {t - 3} \right)}} + {{\rm{e}}^{2\left( {4 - t} \right)}}} $.
We find the critical point of $f\left( t \right)$ by solving the equation $f'\left( t \right) = 0$:
$f'\left( t \right) = \frac{{2{{\rm{e}}^{2\left( {t - 3} \right)}} - 2{{\rm{e}}^{2\left( {4 - t} \right)}}}}{{2\sqrt {1 + {{\rm{e}}^{2\left( {t - 3} \right)}} + {{\rm{e}}^{2\left( {4 - t} \right)}}} }} = \frac{{{{\rm{e}}^{2\left( {t - 3} \right)}} - {{\rm{e}}^{2\left( {4 - t} \right)}}}}{{\sqrt {1 + {{\rm{e}}^{2\left( {t - 3} \right)}} + {{\rm{e}}^{2\left( {4 - t} \right)}}} }}$
The equation $f'\left( t \right) = 0$ gives ${{\rm{e}}^{2\left( {t - 3} \right)}} - {{\rm{e}}^{2\left( {4 - t} \right)}} = 0$.
So,
$2\left( {t - 3} \right) = 2\left( {4 - t} \right)$
$4t = 14$, ${\ \ \ }$ $t = \frac{7}{2}$
The solution is $t = \frac{7}{2}$.
Since the speed function is ever increasing for $t > 0$, the speed at $t = \frac{7}{2}$ is minimum.
Substituting $t = \frac{7}{2}$ in $||{\bf{r}}'\left( t \right)||$ we obtain the minimum speed:
$||{\bf{r}}'\left( {\frac{7}{2}} \right)|| = \sqrt {1 + {{\rm{e}}^{2\left( {\frac{7}{2} - 3} \right)}} + {{\rm{e}}^{2\left( {4 - \frac{7}{2}} \right)}}} = \sqrt {1 + 2{\rm{e}}} $
