Question

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

by the mean value theorem, we know there exists a \\(c\\) in the open interval \\((-5, 3)\\) such that \\(f(c)\\) is equal to this mean slope. for this problem, there is only one \\(c\\) that works. find it.

Explanation:

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<reasoning>

Calculate the function values at the interval endpoints

\[

$$\begin{aligned} f(3) &= 5 - 5(3)^2 = 5 - 45 = -40 \\ f(-5) &= 5 - 5(-5)^2 = 5 - 125 = -120 \end{aligned}$$

\]

Compute the average slope on the interval

\[
\frac{f(3) - f(-5)}{3 - (-5)} = \frac{-40 - (-120)}{3 + 5} = \frac{80}{8} = 10
\]

Find the value of c satisfying the Mean Value Theorem

\[

$$\begin{aligned} f'(x) &= -10x \\ f'(c) &= 10 \implies -10c = 10 \implies c = -1 \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

Consider the function \(f(x) = 5 - 5x^2\) on the interval \([-5, 3]\). Find the average or mean slope of the function on this interval, i.e.
\(\frac{f(3) - f(-5)}{3 - (-5)} =\) <blank>10</blank>

Question 2

By the Mean Value Theorem, we know there exists a \(c\) in the open interval \((-5, 3)\) such that \(f'(c)\) is equal to this mean slope. For this problem, there is only one \(c\) that works. Find it.
<blank>-1</blank>
</answer>

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Answer:

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<reasoning>

Calculate the function values at the interval endpoints

\[

$$\begin{aligned} f(3) &= 5 - 5(3)^2 = 5 - 45 = -40 \\ f(-5) &= 5 - 5(-5)^2 = 5 - 125 = -120 \end{aligned}$$

\]

Compute the average slope on the interval

\[
\frac{f(3) - f(-5)}{3 - (-5)} = \frac{-40 - (-120)}{3 + 5} = \frac{80}{8} = 10
\]

Find the value of c satisfying the Mean Value Theorem

\[

$$\begin{aligned} f'(x) &= -10x \\ f'(c) &= 10 \implies -10c = 10 \implies c = -1 \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

Consider the function \(f(x) = 5 - 5x^2\) on the interval \([-5, 3]\). Find the average or mean slope of the function on this interval, i.e.
\(\frac{f(3) - f(-5)}{3 - (-5)} =\) <blank>10</blank>

Question 2

By the Mean Value Theorem, we know there exists a \(c\) in the open interval \((-5, 3)\) such that \(f'(c)\) is equal to this mean slope. For this problem, there is only one \(c\) that works. Find it.
<blank>-1</blank>
</answer>

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"question_type": "Fill-in-the-blank",
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"Mathematics",
"Calculus",
"Mean Value Theorem"
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