Calculus Question

Question

Using the second derivative, determine where f(x) = x²(1-x) is concave upward? Help!?

Answer 1

I agree with Michael J's conclusion, which should be stated as f"(x) = 2 - 6x = 2(1-3x); it's zero when x=1/3.

Since f"(x) is positive to the left of 1/3 and negative to the right of 1/3, it's true that concavity changes at 1/3 and therefore it's OK to label the point (1/3, 1/3) as an inflection point, and indeed concavity is upward in the interval (-∞, 1/3) and downward the rest of the way.

Answer 2

f(x) = x² - x³

Set second derivative equal to zero.

2x - 3x² = 0

Taking derivative again,

2 - 6x = 0

Solving for x, gives us

x = 1/3

This is the inflection point.  We need to evaluate f'(1/4) and f'(1/2).

f'(1/4) = 2 - (3/2) = 1/2

f'(1/2) = -1

Since f'(1/4) is positive, f(x) is concave up in the interval (-∞, 1/3).