Calculus I - Critical Points

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Section 4.2 : Critical Points

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

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

$g'\left( w \right) = \left( {3{w^2} - 4w - 7} \right){{\bf{e}}^{{w^{\,3}} - 2{w^{\,2}} - 7w}} = \left( {3w - 7} \right)\left( {w + 1} \right){{\bf{e}}^{{w^{\,3}} - 2{w^{\,2}} - 7w}}$

We did some quick factoring to help with the next step.

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Recall that critical points are simply where the derivative is zero and/or doesn’t exist.

This derivative exists everywhere and so we don’t need to worry about that. Therefore, all we need to do is determine where the derivative is zero.

Notice as well that because we know that exponential functions are never zero and so the derivative will only be zero if,

$\left( {3w - 7} \right)\left( {w + 1} \right) = 0\hspace{0.25in} \to \hspace{0.25in}\,w = {\textstyle{7 \over 3}},\,\,\, - 1$

So, we have a two critical points, $\require{bbox} \bbox[2pt,border:1px solid black]{{w = {\textstyle{7 \over 3}}}}$ and $\require{bbox} \bbox[2pt,border:1px solid black]{{w = - 1}}$ for this function.