given the function below
\f(x) = \sqrt3{-98x^3 + 441}\
find the equation of the tangent line to the graph of the function at $x = 1$. answer in $mx + b$ form.
\l(x) =\
use the tangent line to approximate $f(1.1)$.
\l(1.1) =\
compute the actual value of $f(1.1)$. what is the error between the function value and the linear approximation?
answer as a positive value only.
\|\text{error}| \approx\ (approximate to at least 5 decimal places.)

Question

Accepted Answer

{ "quality": "clear", "question_count": 3, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Tangent Line Equation", "Linear Approximation", "Approximation Error" ], "new_concepts": [], "current_concepts": [ "Tangent Line Equation", "Linear Approximation", "Approximation Error" ] } ## Evaluate the function and its derivative at $x = 1$ $$ \begin{aligned} f(x) &= (-98x^3 + 441)^{1/3} \ f(1) &= (-98(1)^3 + 441)^{1/3} = (343)^{1/3} = 7 \ f'(x) &= \frac{1}{3}(-98x^3 + 441)^{-2/3} \cdot (-294x^2) = -98x^2(-98x^3 + 441)^{-2/3} \ f'(1) &= -98(1)^2(-98(1)^3 + 441)^{-2/3} = -98(343)^{-2/3} = -98 \cdot \frac{1}{49} = -2 \end{aligned} $$ ## Determine the tangent line equation and approximate $f(1.1)$ $$ \begin{aligned} L(x) &= f(1) + f'(1)(x - 1) = 7 - 2(x - 1) = -2x + 9 \ L(1.1) &= -2(1.1) + 9 = 6.8 \end{aligned} $$ ## Compute the actual value and the absolute error $$ \begin{aligned} f(1.1) &= (-98(1.1)^3 + 441)^{1/3} = (-98(1.331) + 441)^{1/3} = (310.562)^{1/3} \approx 6.7719335 \ |\text{error}| &= |f(1.1) - L(1.1)| \approx |6.7719335 - 6.8| = 0.0280665 \end{aligned} $$ ## Question 1 $L(x) = -2x + 9$ ## Question 2 $L(1.1) = 6.8$ ## Question 3 $|\text{error}| \approx 0.02807$ { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Calculus", "Linear Approximation" ] }

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Evaluate the function and its derivative at \(x = 1\)

Determine the tangent line equation and approximate \(f(1.1)\)

Compute the actual value and the absolute error

Answer:

Question 1

Question 2

Question 3