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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