Question

What is the local minimum of 4x^3+6x^2-24x+6

Answer 1

and local maximum?

Answer 2

y = = 4x^3 + 3x^2 - 6x + 1

y´ = 12x² + 6x - 6 = 0

6(2x² + x - 1) = 0

(2x - 1)(x + 1) = 0

x = 1/2, -1
y´´ = 24x + 6

y´´(-1) = 24(-1) + 6 = -18 negative, this is a maximum

y´´(1/2) = 24(1/2) + 6 = 18 positive, this is a minimum

y = 4(-1)^3 + 3(-1)^2 - 6(-1) + 1 = 6

y = 4(1/2)^3 + 3(1/2)^2 - 6(1/2) + 1 = -3/4

Maximum.....(-1, 6) <==================
Minimum......(1/2, -3/4) <==================