question 1 (3 points)

use linear approximation at $x = 16$ to estimate $\sqrt{16.4}$.
enter your answer rounded to two decimal places.

Question

question 1 (3 points)

use linear approximation at $x = 16$ to estimate $\sqrt{16.4}$.
enter your answer rounded to two decimal places.

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": [ "Linear Approximation" ], "new_concepts": [], "current_concepts": [ "Linear Approximation" ] } ## Define the function and center point Using the *Linear Approximation* knowledge point $$ \begin{aligned} f(x) &= \sqrt{x} \ a &= 16 \end{aligned} $$ ## Calculate the function value and derivative at the center Using the *Linear Approximation* knowledge point $$ \begin{aligned} f(16) &= \sqrt{16} = 4 \ f'(x) &= \frac{1}{2\sqrt{x}} \ f'(16) &= \frac{1}{2\sqrt{16}} = \frac{1}{8} = 0.125 \end{aligned} $$ ## Construct the linearization formula Using the *Linear Approximation* knowledge point $$ \begin{aligned} L(x) &= f(a) + f'(a)(x - a) \ L(x) &= 4 + 0.125(x - 16) \end{aligned} $$ ## Evaluate the approximation at the target value Using the *Linear Approximation* knowledge point $$ \begin{aligned} x &= 16.4 \ L(16.4) &= 4 + 0.125(16.4 - 16) \ L(16.4) &= 4 + 0.125(0.4) \ L(16.4) &= 4 + 0.05 = 4.05 \end{aligned} $$ Use linear approximation at $x = 16$ to estimate $\sqrt{16.4}$. Enter your answer rounded to two decimal places. 4.05 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Linear Approximation" ] }

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QUESTION IMAGE

Question

Explanation:

Define the function and center point

Calculate the function value and derivative at the center

Construct the linearization formula

Evaluate the approximation at the target value

Answer: