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H(x)=1/8x^3-x^2 what interval has a positive rate of change

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Final answer:

To find the interval where the function H(x) = 1/8x^3 - x^2 has a positive rate of change, we can find the critical points by taking the derivative of the function and solving for x. The interval is (0, 16/3).

Step-by-step explanation:

To determine the interval where the function H(x) = 1/8x^3 - x^2 has a positive rate of change, we can find the critical points of the function by taking its derivative. The derivative of H(x) is given by:

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

To find the critical points, we set H'(x) = 0 and solve for x. We have:

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

This equation can be factored as:

(x)((3/8)x - 2) = 0

Setting each factor equal to zero, we find the critical points:

x = 0 and x = 16/3

Since H'(x) is a quadratic function, it opens upward, meaning that it is positive between the critical points. Therefore, the interval where H(x) has a positive rate of change is:

(0, 16/3)

User Delian Krustev
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