Final answer:
The average rate of change of the polynomial f(x) = x^3 - 9x over the interval (-3, 3) is calculated using the values of the polynomial at the endpoints of the interval and is found to be 0.
Step-by-step explanation:
To calculate the average rate of change of the polynomial f(x) = x^3 - 9x over a given interval, we apply the formula:
\[\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}\]
Let's calculate the average rate of change for each of the provided intervals:
- Interval a) (-1, 0):
\[f(-1) = (-1)^3 - 9(-1) = -1 + 9 = 8\]
\[f(0) = (0)^3 - 9(0) = 0\]
\[\text{Average Rate of Change} = \frac{0 - 8}{0 - (-1)} = \frac{-8}{1} = -8\]
- Interval b) (0, 3):
\[f(3) = (3)^3 - 9(3) = 27 - 27 = 0\]\
[\text{Average Rate of Change} = \frac{0 - 0}{3 - 0} = 0\]
- Interval c) (3, -1):
\[f(3) was already calculated as 0, and f(-1) as 8\]
\[\text{Average Rate of Change} = \frac{8 - 0}{-1 - 3} = \frac{8}{-4} = -2\]
- Interval d) (-3, 3):
\[f(-3) = (-3)^3 - 9(-3) = -27 + 27 = 0\]
\[\text{Average Rate of Change} = \frac{0 - 0}{3 - (-3)} = \frac{0}{6} = 0\]
So the average rate of change of the polynomial on the interval (-3, 3) is 0.