consider the function $f(x) = \frac{x}{x^2 - x + 9}$, $0 \le x \le 11$.

the absolute maximum of $f(x)$ (on the given interval) is at $x = $

and the absolute minimum of $f(x)$ (on the given interval) is at $x = $

Question

consider the function $f(x) = \frac{x}{x^2 - x + 9}$, $0 \le x \le 11$.

the absolute maximum of $f(x)$ (on the given interval) is at $x = $

and the absolute minimum of $f(x)$ (on the given interval) is at $x = $

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Critical Points", "Absolute Extrema on Closed Intervals", "Extreme Value Theorem" ], "new_concepts": [], "current_concepts": [ "Critical Points", "Absolute Extrema on Closed Intervals", "Extreme Value Theorem" ] } ## Find the derivative of the function Using the *Critical Points* knowledge point $$ \begin{aligned} f'(x) &= \frac{1 \cdot (x^2 - x + 9) - x \cdot (2x - 1)}{(x^2 - x + 9)^2} \ f'(x) &= \frac{x^2 - x + 9 - 2x^2 + x}{(x^2 - x + 9)^2} \ f'(x) &= \frac{9 - x^2}{(x^2 - x + 9)^2} \end{aligned} $$ ## Determine the critical points Using the *Critical Points* knowledge point $$ \begin{aligned} 9 - x^2 &= 0 \ x &= 3 \quad (\text{since } -3 otin [0, 11]) \end{aligned} $$ ## Evaluate the function at critical points and endpoints Using the *Absolute Extrema on Closed Intervals* and *Extreme Value Theorem* knowledge points $$ \begin{aligned} f(0) &= \frac{0}{0^2 - 0 + 9} = 0 \ f(3) &= \frac{3}{3^2 - 3 + 9} = \frac{3}{9} = \frac{1}{3} \ f(11) &= \frac{11}{11^2 - 11 + 9} = \frac{11}{119} \approx 0.0924 \end{aligned} $$ ## Identify the absolute maximum and minimum locations Using the *Absolute Extrema on Closed Intervals* and *Extreme Value Theorem* knowledge points $$ \begin{aligned} \text{Maximum value: } &\frac{1}{3} \quad \text{at } x = 3 \ \text{Minimum value: } &0 \quad \text{at } x = 0 \end{aligned} $$ Consider the function $f(x) = \frac{x}{x^2 - x + 9}$, $0 \le x \le 11$. The absolute maximum of $f(x)$ (on the given interval) is at $x =$ 3 and the absolute minimum of $f(x)$ (on the given interval) is at $x =$ 0 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Absolute Extrema on Closed Intervals" ] }

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QUESTION IMAGE

Question

Explanation:

Find the derivative of the function

Determine the critical points

Evaluate the function at critical points and endpoints

Identify the absolute maximum and minimum locations

Answer: