Question

consider the function (f(x) = 5x^3 - 2x) on the interval (-3, 3). find the average or mean slope of the function on this interval.

by the mean value theorem, we know there exists at least one (c) in the open interval ((-3, 3)) 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 \([-3, 3]\)

$$
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Find the derivative \(f'(x)\)

$$ f'(x) = 15x^2 - 2 $$

Solve for \(c\) where \(f'(c) = 43\)

$$
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$$

Answer:

Consider the function \(f(x) = 5x^3 - 2x\) on the interval \([-3, 3]\). Find the average or mean slope of the function on this interval. <blank>43</blank>

By the Mean Value Theorem, we know there exists at least one \(c\) in the open interval \((-3, 3)\) 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>\(-\sqrt{3}\)</blank>
and the larger one is <blank>\(\sqrt{3}\)</blank>