Find and classify all the turning points of y=2x^3+9x^2-24x-1?

1 Answer
Sep 5, 2017

Maximum (-4,111)

Minimum (1,-14)

Explanation:

y = 2x^3 + 9x^2 - 24x - 1

(dy)/(dx) = 6x^2 + 18x - 24

(d^2y)/(dx^2) = 12x + 18

For critical points need (dy)/(dx) = 0.

therefore 0 = 6x^2 + 18x - 24

Divide both sides by 6:

x^2 + 3x - 4 = 0

(x+4)(x-1) = 0

implies x = -4 " or " x = 1

For the nature of these points, need to examine the curvature which is given by the second derivative:

color(red) (x=-4)

(d^2y)/(dx^2)|_(x=-4) = 12(-4) + 18 = -30 < 0

This means that x = -4 is a maximum.

y(-4) = 2(-4)^3 + 9(-4)^2 - 24(-4)-1

y(-4) = -128 + 144 + 96 - 1 = 111

therefore (-4, 111) " is a maximum"

color(red)(x=1)

(d^2y)/(dx^2)|_(x=1) = 12(1) + 18 = 30 > 0

This means that x = 1 is a minimum.

y(1) = 2(1)^3 + 9(1)^2 - 24(1)-1

y(1) = 2 + 9 - 24 - 1 = -14

therefore (1,-14) " is a minimum"