222k views
4 votes
H(x) = 1/8x^3 - x^2

Over which interval does h have a positive rate of change?

A. (-[infinity], -2)
B. (-2, 0)
C. (0, 2)
D. (2, [infinity])

User Monolith
by
7.2k points

1 Answer

6 votes

Final answer:

To determine over which interval the function h(x) = 1/8x^3 - x^2 has a positive rate of change, we need to find the intervals where the derivative of the function is positive. The function h(x) has a positive rate of change over the interval (-∞, 0) U (8/3, ∞).

Step-by-step explanation:

To determine over which interval the function h(x) = 1/8x^3 - x^2 has a positive rate of change, we need to find the intervals where the derivative of the function is positive. Let's find the derivative of h(x):

h'(x) = (3/8)x^2 - 2x

To find where h'(x) > 0, we need to solve the inequality:

(3/8)x^2 - 2x > 0

Simplifying and factoring, we get:

(x - 0)(x - 8/3) > 0

This inequality is true when either both factors are positive or both factors are negative:

x - 0 > 0 and x - 8/3 > 0, or x - 0 < 0 and x - 8/3 < 0

Simplifying, we get:

x > 0 and x > 8/3, or x < 0 and x < 8/3

The solution set is:

x > 8/3 or x < 0

Therefore, the function h(x) has a positive rate of change over the interval (-∞, 0) U (8/3, ∞).

User Whizzzkid
by
7.3k points