use linear approximation, i.e. the tangent line, to approximate $\sqrt{16.3}$. let $f(x) = \sqrt{x}$.

a. find the equation of the tangent line to $f(x)$ at $a = 16$.
$l(x) =$

b. using the linear approximation, find the approximation for $\sqrt{16.3}$.
exact value:
approximate value (to four decimal places):

Question

use linear approximation, i.e. the tangent line, to approximate $\sqrt{16.3}$. let $f(x) = \sqrt{x}$.

a. find the equation of the tangent line to $f(x)$ at $a = 16$.
$l(x) =$

b. using the linear approximation, find the approximation for $\sqrt{16.3}$.
exact value:
approximate value (to four decimal places):

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Linear Approximation", "Tangent Line Equation" ], "new_concepts": [], "current_concepts": [ "Linear Approximation", "Tangent Line Equation" ] } ## Find the tangent line equation at $a = 16$ $$ \begin{aligned} f(x) &= \sqrt{x} \implies f(16) = \sqrt{16} = 4 \ f'(x) &= \frac{1}{2\sqrt{x}} \implies f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8} \ L(x) &= f(16) + f'(16)(x - 16) = 4 + \frac{1}{8}(x - 16) = \frac{1}{8}x + 2 \end{aligned} $$ ## Calculate the exact value of the linear approximation $$ \begin{aligned} L(16.3) &= 4 + \frac{1}{8}(16.3 - 16) \ &= 4 + \frac{1}{8}(0.3) \ &= 4 + \frac{3}{80} \ &= \frac{323}{80} \end{aligned} $$ ## Calculate the approximate value to four decimal places $$ \begin{aligned} L(16.3) &= 4 + 0.0375 \ &= 4.0375 \end{aligned} $$ A. Find the equation of the tangent line to $f(x)$ at $a = 16$. $L(x) =$ $\frac{1}{8}x + 2$ B. Using the linear approximation, find the approximation for $\sqrt{16.3}$. Exact value: $\frac{323}{80}$ Approximate value (to four decimal places): 4.0375 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Linear Approximation" ] }

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Question

Explanation:

Find the tangent line equation at \(a = 16\)

Calculate the exact value of the linear approximation

Calculate the approximate value to four decimal places

Answer: