The first thing we should do is actually verify that Rolle’s Theorem can be used here.

The function is a polynomial which is continuous and differentiable everywhere and so will be continuous on \(\left[ { - 2,1} \right]\) and differentiable on \(\left( { - 2,1} \right)\).

Next, a couple of quick function evaluations shows that \(g\left( { - 2} \right) = g\left( 1 \right) = 0\).

Therefore, the conditions for Rolle’s Theorem are met and so we can actually do the problem.

Note that this may seem to be a little silly to check the conditions but it is a really good idea to get into the habit of doing this stuff. Since we are in this section it is pretty clear that the conditions will be met or we wouldn’t be asking the problem. However, once we get out of this section and you want to use the Theorem the conditions may not be met. If you are in the habit of not checking you could inadvertently use the Theorem on a problem that can’t be used and then get an incorrect answer.

Now that we know that Rolle’s Theorem can be used there really isn’t much to do. All we need to do is take the derivative,

\[g'\left( t \right) = 2 - 2t - 3{t^2}\]

and then solve \(g'\left( c \right) = 0\).

\[ - 3{c^2} - 2c + 2 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}c = \frac{1 \pm \sqrt{7}}{-3} = - 1.2153,\,\,\,0.5486\]

So, we found two values and, in this case, they are both in the interval so the values we want are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{c =\frac{1 \pm \sqrt{7}}{-3} = - 1.2153,\,\,\,0.5486}}\]