do not use a calculator for this problem. find the x-value where the relative maximum of $y = e^{\frac{x}{5}} - x^2$ closest to $x = 0$ occurs. when you need to solve for an expression equal to zero, use newtons method with $x = 0$ as your first guess and enter your second guess as the answer. enter your answer as a fraction.

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{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Derivative of Exponential Functions", "Newtons Method" ], "new_concepts": [], "current_concepts": [ "Derivative of Exponential Functions", "Newtons Method", "Local Extrema Classification" ] } ## Find the derivative of the function $$ \begin{aligned} y &= e^{\frac{x}{5}} - x^2 \ y' &= \frac{1}{5}e^{\frac{x}{5}} - 2x \end{aligned} $$ ## Set up Newton's method for the critical point equation $$ \begin{aligned} f(x) &= y' = \frac{1}{5}e^{\frac{x}{5}} - 2x \ f'(x) &= \frac{1}{25}e^{\frac{x}{5}} - 2 \end{aligned} $$ ## Calculate the second guess using Newton's method $$ \begin{aligned} x_0 &= 0 \ x_1 &= x_0 - \frac{f(x_0)}{f'(x_0)} \ f(0) &= \frac{1}{5}e^0 - 0 = \frac{1}{5} \ f'(0) &= \frac{1}{25}e^0 - 2 = \frac{1}{25} - 2 = -\frac{49}{25} \ x_1 &= 0 - \frac{\frac{1}{5}}{-\frac{49}{25}} = \frac{1}{5} \cdot \frac{25}{49} = \frac{5}{49} \end{aligned} $$ Do not use a calculator for this problem. Find the x-value where the relative maximum of $y = e^{\frac{x}{5}} - x^2$ closest to $x = 0$ occurs. When you need to solve for an expression equal to zero, use Newton's method with $x = 0$ as your first guess and enter your second guess as the answer. Enter your answer as a fraction. $\frac{5}{49}$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Newtons Method" ] }

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Explanation:

Find the derivative of the function

Set up Newton's method for the critical point equation

Calculate the second guess using Newton's method

Answer: