Question

Find the average rate of change of the function over the given intervals. f(x) = 9x3 + 9; [6,8], [-1,1] The average rate of change of the function f(x) = 9x3 + 9 over the interval [6,8] is (Simplify your answer.) The average rate of change of the function f(x) = 9x3 + 9 over the interval [-1,1] is (Simplify your answer.)

Answer 1

Answer:

The average rate of change of the function f(x) = 9x3 + 9 over the interval [6,8] is 1332

The average rate of change of the function f(x) = 9x3 + 9 over the interval [-1,1] is 9.

Step-by-step explanation:

Given a function y, the average rate of change S of $y=f(x)$ in an interval  $(x_{s}, x_{f})$ will be given by the following equation:

$S = \frac{f(x_{f}) - f(x_{s})}{x_{f} - x_{s}}$

In this problem, we have that:

$f(x) = 9x^{3} + 9$

Interval [6,8].

So $x_{s} = 6, x_{f} = 8$

$f(x_{f}) = f(8) = 4617$

$f(x_{s}) = f(6) = 1953$

So

$S = \frac{f(x_{f}) - f(x_{s})}{x_{f} - x_{s}} = \frac{4617 - 1953}{8 - 6} = 1332$

The average rate of change of the function f(x) = 9x3 + 9 over the interval [6,8] is 1332.

Interval [-1,1].

So $x_{s} = -1, x_{f} = 1$

$f(x_{f}) = f(1) = 18$

$f(x_{s}) = f(-1) = 0$

So

$S = \frac{f(x_{f}) - f(x_{s})}{x_{f} - x_{s}} = \frac{18 - 0}{1 - (-1)} = 9$

The average rate of change of the function f(x) = 9x3 + 9 over the interval [-1,1] is 9.