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

Textbook Question

Chapter 4, 4.1 - Problem 32 - 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 + x^2 + 2x)`

`f'(x) = 6x^2 + 2x + 2`

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

`6x^2 + 2x + 2= 0`

You need to divide by 2:

`3x^2 + x + 1 = 0`

Using quadratic formula, yields:

`x_(1,2) = (-1+-sqrt(1 - 4*3))/(2*3) => x_(1,2) = (-1+-sqrt(-11))/6`

Notice that `Delta = -11 < 0` , hence, `x_(1,2) !in R`

Hence, evaluating the critical values of the function yields that there are no real values for f'(x) = 0.

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