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Section 4.4 : Finding Absolute Extrema

5. Determine the absolute extrema of \(h\left( z \right) = 4{z^3} - 3{z^2} + 9z + 12\) on \(\left[ { - 2,1} \right]\).

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Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem!
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First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem!

Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function.

Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section.

Here are the critical points for this function.

\[h'\left( z \right) = 12{z^2} - 6z + 9 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}\,z = \frac{{6 \pm \sqrt { - 396} }}{{24}} = \frac{{1 \pm \sqrt {11} \,i}}{4}\]

Now, recall that we only work with real numbers here and so we ignore complex roots. Therefore, this function has no critical points.

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Technically the next step is to determine all the critical points that are in the given interval. However, there are no critical points for this function and so there are also no critical points in the given interval.

Show Step 3

The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. However, since there are no critical points for this function all we need to do is evaluate the function at the end points of the interval.

Here are those function evaluations.

\[h\left( { - 2} \right) = - 50\hspace{0.5in}h\left( 1 \right) = 22\]

Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem. That is especially true for this problem as there would be no points to evaluate at without the end points.

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The final step is to identify the absolute extrema. So, the answers for this problem are then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{Absolute Maximum : }} & 22 {\mbox{ at }}z = 1\\ {\mbox{Absolute Minimum : }} & - 50{\mbox{ at }}z = - 2\end{align*}}\]

Note that if we hadn’t remembered to evaluate the function at the end points of the interval we would not have had an answer for this problem!