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### Section 4-4 : Finding Absolute Extrema

7. Determine the absolute extrema of $$Q\left( x \right) = {\left( {2 - 8x} \right)^4}{\left( {{x^2} - 9} \right)^3}$$ on $$\left[ { - 3,3} \right]$$.

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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!
Start Solution

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.

\begin{align*}Q'\left( x \right) & = 4\left( { - 8} \right){\left( {2 - 8x} \right)^3}{\left( {{x^2} - 9} \right)^3} + 3\left( {2x} \right){\left( {2 - 8x} \right)^4}{\left( {{x^2} - 9} \right)^2}\\ & = - 4{\left( {2 - 8x} \right)^3}{\left( {{x^2} - 9} \right)^2}\left( {20{x^2} - 3x - 72} \right)\\ & = 0\hspace{0.5in}\hspace{0.25in}\,\, \Rightarrow \hspace{0.25in}x = {\frac{1}{4}},\,\,\,\,x = \pm \,3,\,\,\,\,x = \frac{3 \pm \sqrt {5769}}{40} = - 1.8239,\,\,1.9739\end{align*} Show Step 2

Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, we need all the critical points from the first step.

$x = {\frac{1}{4}},\,\,\,\,x = \pm \,3,\,\,\,\,x = \frac{3 \pm \sqrt {5769}}{40} = - 1.8239,\,\,1.9739$

Do not get excited about the fact that both end points of the interval are also critical points. It happens sometimes and in this case it will reduce the number of computations required in the next step by 2 and that’s not a bad thing.

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. Here are those function evaluations.

$Q\left( { - 3} \right) = 0\,\,\,\,\,\,Q\left( { - 1.8239} \right) = - 1.38 \times {10^7}\,\,\,\,\,Q\left( {{\frac{1}{4}}} \right) = 0\,\,\,\,\,\,Q\left( {1.9739} \right) = - 4.81 \times {10^6}\,\,\,\,\,Q\left( 3 \right) = 0$

Do not get excited about the large numbers for the two non-zero function values. This is something that is going to happen on occasion and we shouldn’t worry about it when it does happen.

Show Step 4

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 : }} & 0 {\mbox{ at }}x = - 3,\,\,\,x = {\frac{1}{4}},\,\,x = 3\\ {\mbox{Absolute Minimum : }} & - 1.38 \times {10^7}{\mbox{ at }}x = - 1.8239\end{align*}}

Recall that while we can only have one largest possible value (i.e. only one absolute maximum) it is completely possible for it to occur at more than one point (3 points in this case).