h(x)=\\frac{1}{8}x^{3}-x^{2}\nover which interval does h have a positive average rate of change?\nchoose 1…

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function (y = h(x)) over the interval ([a,b]) is (\frac{h(b)-h(a)}{b - a}). Since (b-a>0) (because (b>a) for an interval ([a,b])), we just need to check if (h(b)-h(a)>0) (i.e., (h(b)>h(a))).

Step2: Calculate (h(x)) for option A

For (a = 0) and (b=2): (h(0)=\frac{1}{8}(0)^{3}-(0)^{2}=0) (h(2)=\frac{1}{8}(2)^{3}-(2)^{2}=\frac{8}{8}-4=1 - 4=-3) (h(2)-h(0)=-3-0=-3<0)

Step3: Calculate (h(x)) for option B

For (a = 6) and (b = 8): (h(6)=\frac{1}{8}(6)^{3}-(6)^{2}=\frac{216}{8}-36 = 27-36=-9) (h(8)=\frac{1}{8}(8)^{3}-(8)^{2}=\frac{512}{8}-64=64 - 64=0) (h(8)-h(6)=0-(-9)=9>0)

Step4: Calculate (h(x)) for option C

For (a = 0) and (b = 6): (h(0) = 0) (from step 2) (h(6)=-9) (from step 3) (h(6)-h(0)=-9 - 0=-9<0)

Step5: Calculate (h(x)) for option D

For (a = 0) and (b = 8): (h(0)=0) (from step 2) (h(8)=0) (from step 3) (h(8)-h(0)=0-0 = 0)

Answer:

B. (6\leq x\leq8)