Question

H(x)= 1/8x^3−x^2

Over which interval does h(x) have a positive average rate of change?
a) 0≤x≤2
b) 0≤x≤6
c) 6≤x≤8
d) 0≤x≤8

Answer 1

The interval over which the function h(x) = 1/8x^3 - x^2 has a positive average rate of change is 0 ≤ x ≤ 2.

Step-by-step explanation:

To determine the interval over which the function h(x) = 1/8x^3 - x^2 has a positive average rate of change, we need to find the values of x for which the function's derivative is positive. The derivative of h(x) is h'(x) = 3/8x^2 - 2x. Setting h'(x) > 0, we can solve for x to find the interval:

3/8x^2 - 2x > 0.
x(3/8x - 2) > 0.
This inequality holds true when either both factors are positive or both factors are negative. By solving each of these cases separately, we find that the interval over which h(x) has a positive average rate of change is 0 ≤ x ≤ 2.