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Section 4.2 : Critical Points

4. Determine the critical points of \(g\left( x \right) = {x^6} - 2{x^5} + 8{x^4}\).

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

\[g'\left( x \right) = 6{x^5} - 10{x^4} + 32{x^3} = 2{x^3}\left( {3{x^2} - 5x + 16} \right)\]

Factoring the derivative as much as possible will 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. 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.

\[2{x^3}\left( {3{x^2} - 5x + 16} \right) = 0\hspace{0.5in} \Rightarrow \hspace{0.25in}2{x^3} = 0\hspace{0.5in}{\mbox{OR}}\hspace{0.25in}3{x^2} - 5x + 16 = 0\]

From the first term we clearly see that \(x = 0\) is a critical point. The second term does not factor and we we’ll need to use the quadratic formula to solve this equation.

\[x = \frac{{5 \pm \sqrt { - 167} }}{6} = \frac{{5 \pm \sqrt {167} \,i}}{6}\]

Remember that not all quadratics will factor so don’t forget about the quadratic formula!

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Now, recall that we don’t use complex numbers in this class and so the solutions from the second term are not critical points. Therefore, the only critical point of this function is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0}}\]