`f(x) = 2x^3 - 3x^2 - 36x` Find the critical numbers of the function

Textbook Question

Chapter 4, 4.1 - Problem 31 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the critical numbers of the function, hence, you need to evaluate the soutions to the first derivative, such that:

`f'(x) = 0`

`f'(x) = (2x^3 - 3x^2 - 36x)`

`f'(x) = 6x^2 - 6x - 36`

You need to solve for x the equation f'(x) = 0:

`6x^2 - 6x - 36= 0`

You need to divide by 6:

`x^2 - x - 6 = 0`

Using quadratic formula, yields:

`x_(1,2) = (1+-sqrt(1 + 4*6))/2 => x_(1,2) = (1+-sqrt(25))/2`

`x_(1,2) = (1+-5)/2 => x_1 = 3 ; x_2 = -2`

Hence, evaluating the critical numbers of the function yields (3,f(3)), (-2,f(-2)).

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