the function $f(x) = 3 + 4x + 36x^{-1}$ has one local minimum and one local maximum.
this function has a local maximum at $x =$
with value
and a local minimum at $x =$
with value

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Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Critical Points", "First Derivative Test", "Local Extrema Classification" ], "new_concepts": [], "current_concepts": [ "Critical Points", "First Derivative Test", "Local Extrema Classification" ] } ## Find the first derivative and critical points $$ \begin{aligned} f(x) &= 3 + 4x + 36x^{-1} \ f'(x) &= 4 - 36x^{-2} = 4 - \frac{36}{x^2} \ 4 - \frac{36}{x^2} &= 0 \implies x^2 = 9 \implies x = \pm 3 \end{aligned} $$ ## Classify the critical points using the second derivative $$ \begin{aligned} f''(x) &= 72x^{-3} = \frac{72}{x^3} \ f''(3) &= \frac{72}{27} > 0 \implies \text{local minimum at } x = 3 \ f''(-3) &= \frac{72}{-27} < 0 \implies \text{local maximum at } x = -3 \end{aligned} $$ ## Calculate the local extremum values $$ \begin{aligned} f(-3) &= 3 + 4(-3) + \frac{36}{-3} = 3 - 12 - 12 = -21 \ f(3) &= 3 + 4(3) + \frac{36}{3} = 3 + 12 + 12 = 27 \end{aligned} $$ The function $f(x) = 3 + 4x + 36x^{-1}$ has one local minimum and one local maximum. This function has a local maximum at $x =$ -3 with value -21 and a local minimum at $x =$ 3 with value 27 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Local Extrema Classification" ] }

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Question

Explanation:

Find the first derivative and critical points

Classify the critical points using the second derivative

Calculate the local extremum values

Answer: