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### Section 4.2 : Critical Points

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

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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.

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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.