We know that the derivative of the function will give us the rate of change for the function and so we’ll need that.

\[V'\left( t \right) = \frac{{\left( 1 \right){{\bf{e}}^t} - t\left( {{{\bf{e}}^t}} \right)}}{{{{\left( {{{\bf{e}}^t}} \right)}^2}}} = \frac{{{{\bf{e}}^t} - t{{\bf{e}}^t}}}{{{{\left( {{{\bf{e}}^t}} \right)}^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1 - t}}{{{{\bf{e}}^t}}}}}\]

Now, all we need to do is evaluate the derivative at the point in question. So,

\[V'\left( { - 4} \right) = \frac{5}{{{{\bf{e}}^{ - 4}}}} = 272.991 > 0\]

\(V'\left( { - 4} \right) > 0\) and so the function must be increasing at \(t = - 4\).

We found the derivative of the function in the first part so here all we need to do is the evaluation.

\[V'\left( 0 \right) = \frac{1}{{{{\bf{e}}^0}}} = 1 > 0\]

\(V'\left( 0 \right) > 0\) and so the function must be increasing at \(t = 0\).

We found the derivative of the function in the first part so here all we need to do is the evaluation.

\[V'\left( {10} \right) = \frac{{ - 9}}{{{{\bf{e}}^{10}}}} = - 0.0004086 < 0\]

\(V'\left( {10} \right) < 0\) and so the function must be decreasing at \(t = 10\).