Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Applications of Derivatives / Critical Points
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 4-2 : Critical Points

6. Determine the critical points of \(Q\left( x \right) = {\left( {2 - 8x} \right)^4}{\left( {{x^2} - 9} \right)^3}\).

Show All Steps Hide All Steps

Start Solution

We’ll need the first derivative to get the answer to this problem so let’s get that.

\[\begin{align*}Q'\left( x \right) & = 4{\left( {2 - 8x} \right)^3}\left( { - 8} \right){\left( {{x^2} - 9} \right)^3} + {\left( {2 - 8x} \right)^4}\left( 3 \right){\left( {{x^2} - 9} \right)^2}\left( {2x} \right)\\ & = 2{\left( {2 - 8x} \right)^3}{\left( {{x^2} - 9} \right)^2}\left[ { - 16\left( {{x^2} - 9} \right) + 3x\left( {2 - 8x} \right)} \right]\\ & = 2{\left( {2 - 8x} \right)^3}{\left( {{x^2} - 9} \right)^2}\left[ { - 40{x^2} + 6x + 144} \right] = - 4{\left( {2 - 8x} \right)^3}{\left( {{x^2} - 9} \right)^2}\left[ {20{x^2} - 3x - 72} \right]\end{align*}\]

Factoring the derivative as much as possible will help with the next step. For this problem (unlike some of the previous problems) this extra factoring is all but required to make this easier to finish.

Show Step 2

Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial, (admittedly a somewhat messy polynomial) and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points.

\[ - 4{\left( {2 - 8x} \right)^3}{\left( {{x^2} - 9} \right)^2}\left[ {20{x^2} - 3x - 72} \right] = 0\]

From this we get the following three equations that we need to solve.

\[\begin{align*}{\left( {2 - 8x} \right)^3} & = 0\\ {\left( {{x^2} - 9} \right)^2} & = 0\\ 20{x^2} - 3x - 72 & = 0\end{align*}\]

For the first two equations all we really need to do is set the quantity inside the parenthesis to zero (the exponent on the parenthesis won’t affect the solution) and the third requires the quadratic formula.

\[2 - 8x = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{x = {\textstyle{1 \over 4}}}}\] \[{x^2} - 9 = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{x = \pm 3}}\] \[20{x^2} - 3x - 72 = 0\hspace{0.5in}\Rightarrow \hspace{0.25in}\,\,x = \frac{{3 \pm \sqrt {{3^2} - 4\left( {20} \right)\left( { - 72} \right)} }}{{2\left( {20} \right)}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{3 \pm \sqrt {5769} }}{{40}}}}\]

So, we get the 5 critical points boxed in above.