Question

consider the function (f(x) = \frac{1}{x}) on the interval (2, 6). 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 ((2, 6)) such that (f(c)) is equal to this mean slope. for this problem, there is only one (c) that works. find it.

Explanation:

<pre_analysis>
{
"quality": "clear",
"question_count": 2,
"skills_matched": [
"step_cot",
"explain_then_answer",
"quick"
],
"needs_drawing": false,
"concepts_used": [
"Average Rate of Change",
"Mean Value Theorem"
],
"new_concepts": [],
"current_concepts": [
"Average Rate of Change",
"Mean Value Theorem",
"Derivative Definition"
]
}
</pre_analysis>

<reasoning>

Calculate the average slope on the interval

\[

$$\begin{aligned} &a = 2,\quad b = 6\\ &f(2) = \frac{1}{2},\quad f(6) = \frac{1}{6}\\ &\text{Average Slope} = \frac{f(6) - f(2)}{6 - 2} = \frac{\frac{1}{6} - \frac{1}{2}}{4} = \frac{-\frac{1}{3}}{4} = -\frac{1}{12} \end{aligned}$$

\]

Find the derivative of the function

\[

$$\begin{aligned} &f(x) = x^{-1}\\ &f'(x) = -x^{-2} = -\frac{1}{x^2} \end{aligned}$$

\]

Solve for c using the Mean Value Theorem

\[

$$\begin{aligned} &f'(c) = -\frac{1}{12}\\ &-\frac{1}{c^2} = -\frac{1}{12}\\ &c^2 = 12\\ &c = \sqrt{12} = 2\sqrt{3} \approx 3.4641 \quad (\text{since } c \in (2, 6)) \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

\(-\frac{1}{12}\)

Question 2

\(2\sqrt{3}\)
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Calculus",
"Mean Value Theorem"
]
}
</post_analysis>

Answer:

<pre_analysis>
{
"quality": "clear",
"question_count": 2,
"skills_matched": [
"step_cot",
"explain_then_answer",
"quick"
],
"needs_drawing": false,
"concepts_used": [
"Average Rate of Change",
"Mean Value Theorem"
],
"new_concepts": [],
"current_concepts": [
"Average Rate of Change",
"Mean Value Theorem",
"Derivative Definition"
]
}
</pre_analysis>

<reasoning>

Calculate the average slope on the interval

\[

$$\begin{aligned} &a = 2,\quad b = 6\\ &f(2) = \frac{1}{2},\quad f(6) = \frac{1}{6}\\ &\text{Average Slope} = \frac{f(6) - f(2)}{6 - 2} = \frac{\frac{1}{6} - \frac{1}{2}}{4} = \frac{-\frac{1}{3}}{4} = -\frac{1}{12} \end{aligned}$$

\]

Find the derivative of the function

\[

$$\begin{aligned} &f(x) = x^{-1}\\ &f'(x) = -x^{-2} = -\frac{1}{x^2} \end{aligned}$$

\]

Solve for c using the Mean Value Theorem

\[

$$\begin{aligned} &f'(c) = -\frac{1}{12}\\ &-\frac{1}{c^2} = -\frac{1}{12}\\ &c^2 = 12\\ &c = \sqrt{12} = 2\sqrt{3} \approx 3.4641 \quad (\text{since } c \in (2, 6)) \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

\(-\frac{1}{12}\)

Question 2

\(2\sqrt{3}\)
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Calculus",
"Mean Value Theorem"
]
}
</post_analysis>