unit 6: quiz: related rates and application of derivatives
question 14
find the critical point(s) within the domain of the following function. answer exactly.
\y = 25x^3 - 12x\
\x = \

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{ "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" ], "new_concepts": [], "current_concepts": [ "Critical Points", "Power Rule", "Solving Quadratic Equations" ] } ## Find the derivative of the function Using the *Critical Points* knowledge point $$ \begin{aligned} y &= 25x^3 - 12x \ y' &= 75x^2 - 12 \end{aligned} $$ ## Set the derivative equal to zero Using the *Critical Points* knowledge point $$ \begin{aligned} 75x^2 - 12 &= 0 \ 75x^2 &= 12 \ x^2 &= \frac{12}{75} \ x^2 &= \frac{4}{25} \ x &= \pm\frac{2}{5} \end{aligned} $$ Find the critical point(s) within the domain of the following function. Answer exactly. $y = 25x^3 - 12x$ $x =$ $\pm\frac{2}{5}$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Critical Points" ] }

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Question

Explanation:

Find the derivative of the function

Set the derivative equal to zero

Answer: