Paul's Online Notes
Home / Calculus I / Applications of Derivatives / Finding Absolute Extrema
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4-4 : Finding Absolute Extrema

8. Determine the absolute extrema of $$h\left( w \right) = 2{w^3}{\left( {w + 2} \right)^5}$$ on $$\left[ { - {\displaystyle \frac{5}{2}},{\displaystyle \frac{1}{2}}} \right]$$.

Show All Steps Hide All Steps

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*}h'\left( w \right) & 6{w^2}{\left( {w + 2} \right)^5} + 10{w^3}{\left( {w + 2} \right)^4}\\ & = 4{w^2}{\left( {w + 2} \right)^4}\left( {4w + 3} \right) = 0\,\hspace{0.5in} \Rightarrow \hspace{0.5in}w = 0,\,\,\,w = - {\frac{3}{4}},\,\,\,w = - 2\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.

$w = 0,\,\,\,w = - {\frac{3}{4}},\,\,\,w = - 2$ 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.

$h\left( { - {\frac{5}{2}}} \right) = 0.9766\,\,\,\,\,\,h\left( { - 2} \right) = 0\,\,\,\,\,\,h\left( { - {\frac{3}{4}}} \right) = - 2.5749\,\,\,\,\,\,\,h\left( 0 \right) = 0\,\,\,\,\,\,\,h\left( {{\frac{1}{2}}} \right) = 24.4141$ 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 : }} & 24.4141{\mbox{ at }}w = {\frac{1}{2}}\\ {\mbox{Absolute Minimum : }}& - 2.5749{\mbox{ at }}w = - {\frac{3}{4}}\end{align*}}