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.

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.

Accepted Answer

{ "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" ] } ## 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 $$ \begin{aligned} &f'(c) = -\frac{1}{12}\ &-\frac{1}{c^2} = -\frac{1}{12}\ &c^2 = 12\ &c = \sqrt{12} = 2\sqrt{3} \approx 3.4641 \quad (\text{since } c \in (2, 6)) \end{aligned} $$ ## Question 1 $-\frac{1}{12}$ ## Question 2 $2\sqrt{3}$ { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Calculus", "Mean Value Theorem" ] }

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Question

Explanation:

Calculate the average slope on the interval

Find the derivative of the function

Solve for c using the Mean Value Theorem

Answer:

Question 1

Question 2