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

h(x) = \\frac{1}{8}x^{3}-x^{2}\nover which interval does h have a positive average rate of change?\nchoose 1 answer:\na 0 \\leq x \\leq 2\nb 0 \\leq x \\leq 8\nc 6 \\leq x \\leq 8\nd 0 \\leq x \\leq 6

h(x) = \\frac{1}{8}x^{3}-x^{2}\nover which interval does h have a positive average rate of change?\nchoose 1 answer:\na 0 \\leq x \\leq 2\nb 0 \\leq x \\leq 8\nc 6 \\leq x \\leq 8\nd 0 \\leq x \\leq 6

Answer

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 given by (\frac{h(b)-h(a)}{b - a}).

Step2: Calculate (h(x)) at the endpoints for each interval

  • For interval (0\leq x\leq2):
    • (h(0)=\frac{1}{8}(0)^{3}-(0)^{2}=0).
    • (h(2)=\frac{1}{8}(2)^{3}-(2)^{2}=\frac{1}{8}\times8 - 4=1 - 4=-3).
    • The average rate of change is (\frac{h(2)-h(0)}{2 - 0}=\frac{-3-0}{2}=-\frac{3}{2}<0).
  • For interval (0\leq x\leq8):
    • (h(0) = 0).
    • (h(8)=\frac{1}{8}(8)^{3}-(8)^{2}=\frac{1}{8}\times512-64 = 64 - 64=0).
    • The average rate of change is (\frac{h(8)-h(0)}{8 - 0}=\frac{0 - 0}{8}=0).
  • For interval (6\leq x\leq8):
    • (h(6)=\frac{1}{8}(6)^{3}-(6)^{2}=\frac{1}{8}\times216-36=\frac{27}{1}-36=-9).
    • (h(8)=0).
    • The average rate of change is (\frac{h(8)-h(6)}{8 - 6}=\frac{0-(-9)}{2}=\frac{9}{2}>0).
  • For interval (0\leq x\leq6):
    • (h(0) = 0).
    • (h(6)=-9).
    • The average rate of change is (\frac{h(6)-h(0)}{6 - 0}=\frac{-9-0}{6}=-\frac{3}{2}<0).

Answer:

C. (6\leq x\leq8)