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:

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<reasoning>

Calculate the average slope on the interval \([-3, 3]\)

\[

$$\begin{aligned} f(x) &= 5x^3 - 2x \\ f(3) &= 5(3)^3 - 2(3) = 135 - 6 = 129 \\ f(-3) &= 5(-3)^3 - 2(-3) = -135 + 6 = -129 \\ \text{Average Slope} &= \frac{f(3) - f(-3)}{3 - (-3)} = \frac{129 - (-129)}{6} = \frac{258}{6} = 43 \end{aligned}$$

\]

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

\[
f'(x) = 15x^2 - 2
\]

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

\[

$$\begin{aligned} 15c^2 - 2 &= 43 \\ 15c^2 &= 45 \\ c^2 &= 3 \\ c &= \pm\sqrt{3} \approx \pm 1.732 \end{aligned}$$

\]
</reasoning>

<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>
</answer>

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Answer:

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<reasoning>

Calculate the average slope on the interval \([-3, 3]\)

\[

$$\begin{aligned} f(x) &= 5x^3 - 2x \\ f(3) &= 5(3)^3 - 2(3) = 135 - 6 = 129 \\ f(-3) &= 5(-3)^3 - 2(-3) = -135 + 6 = -129 \\ \text{Average Slope} &= \frac{f(3) - f(-3)}{3 - (-3)} = \frac{129 - (-129)}{6} = \frac{258}{6} = 43 \end{aligned}$$

\]

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

\[
f'(x) = 15x^2 - 2
\]

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

\[

$$\begin{aligned} 15c^2 - 2 &= 43 \\ 15c^2 &= 45 \\ c^2 &= 3 \\ c &= \pm\sqrt{3} \approx \pm 1.732 \end{aligned}$$

\]
</reasoning>

<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>
</answer>

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