Tangent Lines and Intervals

Question

Determine the points of inflection (if any) of and the open intervals on which the graph of ℎ(x) = 2x^3 − 6x^2 is concave upward or downward by hand.

Answer 1

h(x) = 2x³ - 6x²

h'(x) = 6x² - 12x = 6x(x - 2)

h''(x) = 12x - 12 = 12(x - 1)

Critical points are when h'(x) = 0 so 6x(x - 2) = 0 when x = 0 and x = 2

First Derivative Test:

Since slopes are positive to the left of 0 and negative to the right of 0, there is a local max @ x = 0

Since slopes are negative to the left of 2 and positive to the right of 2, there is a local min @ x = 2

Second Derivative Test:

The second derivative 12(x - 1) = 0 when x = 1

The second derivative is negative to the left of 1 meaning the curve is concave down on (-∞, 1)

The second derivative is positive to the right of 1 meaning the curve is concave up on (1, ∞)

and there is a point of infliction at x = 1 (the point is (1, -4)

Answer 2

ℎ(x) = 2x^3 − 6x^2

h'(x) = 6x^2 - 12x = 0, x = 0 and 2

h''(x) = 12x - 12, x = 1

H(1) = -4

Point of Inflection is (1,-4)

Slope at x=1 is -6, so with a minimum at (2,-6), where the slope turns positive,

The h(x) turns downward at x=0, with the point of inflection at (1,-4) and upwards at x=6.