Question

h(x) = \frac{1}{8}x^{3}-x^{2}
over which interval does h have a positive average rate of change?
choose 1 answer:
a 0 \leq x \leq 2
b 0 \leq x \leq 8
c 6 \leq x \leq 8
d 0 \leq x \leq 6

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. \(6\leq x\leq8\)