Question

consider the function (f(x) = 2x^3 - 6x^2 - 48x + 4) on the interval (-4, 10). find the average or mean slope of the function on this interval.

by the mean value theorem, we know there exists a (c) in the open interval ((-4, 10)) such that (f(c)) is equal to this mean slope. for this problem, there are two values of (c) that work.

the smaller one is

and the larger one is

Explanation:

Calculate the average slope on the interval

$$
LATEXBLOCK0
$$

Find the derivative of the function

$$
LATEXBLOCK1
$$

Solve for c where the derivative equals the average slope

$$
LATEXBLOCK2
$$

Answer:

Consider the function \(f(x) = 2x^3 - 6x^2 - 48x + 4\) on the interval \([-4, 10]\). Find the average or mean slope of the function on this interval.
<blank>68</blank>

By the Mean Value Theorem, we know there exists a \(c\) in the open interval \((-4, 10)\) such that \(f'(c)\) is equal to this mean slope. For this problem, there are two values of \(c\) that work.

The smaller one is <blank>\(1 - \frac{\sqrt{183}}{3}\)</blank>

and the larger one is <blank>\(1 + \frac{\sqrt{183}}{3}\)</blank>