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 6\leq x\leq 8
c 0\leq x\leq 6
d 0\leq x\leq 8

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. \(6\leq x\leq8\)