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Section 4-4 : Finding Absolute Extrema

9. Determine the absolute extrema of \(\displaystyle f\left( z \right) = \frac{{z + 4}}{{2{z^2} + z + 8}}\) on \(\left[ { - 10,0} \right]\).

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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 rational expression in which both the numerator and denominator are continuous everywhere. Also notice that the rational expression exists at all points in the interval 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.

\[\begin{align*}f'\left( z \right) & = \frac{{\left( 1 \right)\left( {2{z^2} + z + 8} \right) - \left( {z + 4} \right)\left( {4z + 1} \right)}}{{{{\left( {2{z^2} + z + 8} \right)}^2}}}\\ & = \frac{{ - 2\left( {{z^2} + 8z - 2} \right)}}{{{{\left( {2{z^2} + z + 8} \right)}^2}}} = 0\,\,\,\,\,\,\,\hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,z = \frac{- 8 \pm \sqrt {72}}{2} = - 4 \pm 3\sqrt 2 = - 8.2426,\,\,\,0.2426\end{align*}\] 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, the only critical point that we need is,

\[z = - 4 - 3\sqrt 2 = - 8.2426\] Show Step 3

The next step is to evaluate the function at the critical point from the second step and at the end points of the given interval. Here are those function evaluations.

\[f\left( { - 10} \right) = - {\frac{1}{33}} = - 0.0303\hspace{0.5in}f\left( { - 8.2426} \right) = - 0.03128\hspace{0.5in}f\left( 0 \right) = {\frac{1}{2}}\] Show Step 4

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{Absolute Maximum : }} & {\frac{1}{2}}{\mbox{ at }}z = 0\\ {\mbox{Absolute Minimum : }} & - 0.03128{\mbox{ at }}z = - 4 - 3\sqrt 2 \end{align*}}\]

Note the importance of paying attention to the interval with this problem. Had we neglected to exclude \(z = - 4 + 3\sqrt 2 = 0.2426\) we would have gotten the wrong answer for the absolute maximum.

This problem also shows that we need to be very careful with doing too much rounding of our answers. Had we rounded down to say 2 decimal places we would have been tempted to say that the absolute minimum occurred at two places when in fact one of the points was lower than the other.