h(x)=\\frac{1}{8}x^{3}-x^{2}\nwhat is the average rate of change of h over the interval -2\\leq x\\leq2?
h(x)=\\frac{1}{8}x^{3}-x^{2}\nwhat is the average rate of change of h over the interval -2\\leq x\\leq2?
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 (\frac{h(b)-h(a)}{b - a}). Here, (a=-2), (b = 2).
Step2: Calculate (h(-2))
Substitute (x=-2) into (h(x)=\frac{1}{8}x^{3}-x^{2}). [ \begin{align*} h(-2)&=\frac{1}{8}\times(-2)^{3}-(-2)^{2}\ &=\frac{1}{8}\times(-8)-4\ &=-1 - 4\ &=-5 \end{align*} ]
Step3: Calculate (h(2))
Substitute (x = 2) into (h(x)=\frac{1}{8}x^{3}-x^{2}). [ \begin{align*} h(2)&=\frac{1}{8}\times2^{3}-2^{2}\ &=\frac{1}{8}\times8 - 4\ &=1-4\ &=-3 \end{align*} ]
Step4: Calculate the average rate of change
Substitute (h(-2)=-5), (h(2)=-3), (a=-2), (b = 2) into (\frac{h(b)-h(a)}{b - a}). [ \begin{align*} \frac{h(2)-h(-2)}{2-(-2)}&=\frac{-3-(-5)}{2 + 2}\ &=\frac{-3 + 5}{4}\ &=\frac{2}{4}\ &=\frac{1}{2} \end{align*} ]
Answer:
(\frac{1}{2})