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### Section 4.2 : Critical Points

3. Determine the critical points of $$g\left( w \right) = 2{w^3} - 7{w^2} - 3w - 2$$.

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

$g'\left( w \right) = 6{w^2} - 14w - 3$ 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 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.

$6{w^2} - 14w - 3 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{w = \frac{{14 \pm \sqrt {268} }}{{12}} = \frac{{7 \pm \sqrt {67} }}{6}}}$

As we can see in this case we needed to use the quadratic formula to find the critical points. Not all quadratics will factor so don’t forget about the quadratic formula!