Find critical values tell if there is a maximum

Question

Find the critical values, tell if there is a maximum y=x^3-9x-6

Answer 1

y = x^3 - 9x - 6
y' = 3x^2 - 9

To find the critical values, we will set y' to zero. We have:

y' = 0
3x^2 - 9 = 0
3(x^2 - 3) = 0
3(x + sqrt(3))(x - sqrt(3)) = 0
(x + sqrt(3)) = 0 or (x - sqrt(3)) = 0
x = -sqrt(3) or x = sqrt(3)

y'' = 6x

When x = -sqrt(3), y'' = 6(-sqrt(3)) = -6sqrt(3) < 0
Thus, we have a local maximum at x = -sqrt(3)

When x = sqrt(3), y'' = 6(sqrt(3)) = 6sqrt(3) > 0
Thus, we have a local minimum at x - sqrt(3).