How do you find the critical points of $k'(x)=x^3-3x^2-18x+40$ ?

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How do you find the critical points of $k'(x)=x^3-3x^2-18x+40$ ?

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Answer 1 · 2018-05-12T18:11:43

$x_1 = 1- sqrt(7), x_2 = 1+ sqrt(7)$

Explanation:

If you want the critical points of $k'(x)$ , you have to compute $k''(x)$ . An easy calculation gives us $k''(x) = 3x^2 - 6x - 18.$ You only have to solve $3x^2 - 6x - 18=0.$

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How do you find the critical points of $k'(x)=x^3-3x^2-18x+40$ ?

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How do you find the critical points of k'(x)=x^3-3x^2-18x+40k'(x)=x^3-3x^2-18x+40?

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Explanation:

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How do you find the critical points of $k'(x)=x^3-3x^2-18x+40$ ?