consider the function $f(x) = 5x^2 - 4x + 5$, $0 \le x \le 6$.

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 $f(x) = 5x^2 - 4x + 5$, $0 \le x \le 6$.

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", "quick" ], "needs_drawing": false, "concepts_used": [ "Critical Points", "Absolute Extrema on Closed Intervals" ], "new_concepts": [], "current_concepts": [ "Critical Points", "Absolute Extrema on Closed Intervals" ] } ## Find critical points of the function $$ \begin{aligned} f(x) &= 5x^2 - 4x + 5, \quad 0 \le x \le 6 \ f'(x) &= 10x - 4 \ 10x - 4 &= 0 \implies x = 0.4 \end{aligned} $$ ## Evaluate the function at critical points and endpoints $$ \begin{aligned} f(0) &= 5(0)^2 - 4(0) + 5 = 5 \ f(0.4) &= 5(0.4)^2 - 4(0.4) + 5 = 0.8 - 1.6 + 5 = 4.2 \ f(6) &= 5(6)^2 - 4(6) + 5 = 180 - 24 + 5 = 161 \end{aligned} $$ ## Identify absolute extrema locations $$ \begin{aligned} \text{Absolute maximum value is } 161 \text{ at } x &= 6 \ \text{Absolute minimum value is } 4.2 \text{ at } x &= 0.4 \end{aligned} $$ Consider the function $f(x) = 5x^2 - 4x + 5$, $0 \le x \le 6$. The absolute maximum of $f(x)$ (on the given interval) is at $x =$ 6 and the absolute minimum of $f(x)$ (on the given interval) is at $x =$ 0.4 { "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 critical points of the function

Evaluate the function at critical points and endpoints

Identify absolute extrema locations

Answer: