Question

H(x)=1/8x^3-x^2

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

Answer 1

Answer: 6 < x < 8

Explanation:

Answer 2

`(-\infty,0)\cup((16)/(3),\infty)\\That \ is x<0 \ or \ x>(16)/(3)`

Explanation:

`h(x)=(1)/(8)x^(3) -x^(2) \\\\Differentiate \ h(x) \ with \  respect \ to \ x\\h'(x)=(3)/(8)x^(2) -2x`

For positive rate of change
`h'(x)>0`

`(3)/(8) x^(2) -2x>0\\x((3)/(8)x-2)>0\\\\ When \ x<0 \Rightarrow ((3)/(8)x-2)<0 \Rightarrow h'(x)>0 \ (multiplication \ of \ two \ negative \ gives \ positive)\\\\When ((3)/(8)x-2)>0 \Rightarrow x>(16)/(3) \Rightarrow x>0\\\\h'(x)=x((3)/(8)x-2) >0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x<0 \ or \ x>(16)/(3) \\\\`