What is the equation of the normal line of f(x)=(x−2)32−x3 at x=2?

Explanation:the given: f(x)=(x−2)32−x3 and x=2f(2)=(2−2)32−(2)3f(2)=-8the point on the curve : (2, -8)compute slope m then use −1m for the normal linef′(x)=(32)(x−2)12−3x2f′(2)=(32)(2−2)12−3(2)2f'(2)=-12The slope to be used to find the normal line is −1m−1m=−1−12=112Determine now the Normal Line usingy−y1=(112)(x−x1)y−(−8)=(112)(x−2)after simplification the final answer:x-12y=98 or y=x12−496

Step by step solution to "What is the equation of the normal line..."

High School
Calculus and Analysis
Calculus 1
Derivatives

Malik Dean

Answered question

2022-04-10

What is the equation of the normal line of

f (x) = {(x - 2)}^{\frac{3}{2}} - x^{3}

at x=2?

Answer & Explanation

ze2m1ingkdvu

Beginner2022-04-11Added 16 answers

Explanation:
the given:

f (x) = {(x - 2)}^{\frac{3}{2}} - x^{3}

and x=2

f (2) = {(2 - 2)}^{\frac{3}{2}} - {(2)}^{3}

f(2)=-8
the point on the curve : (2, -8)
compute slope m then use

- \frac{1}{m}

for the normal line

f^{'} (x) = (\frac{3}{2}) {(x - 2)}^{\frac{1}{2}} - 3 x^{2}

f^{'} (2) = (\frac{3}{2}) {(2 - 2)}^{\frac{1}{2}} - 3 {(2)}^{2}

f'(2)=-12
The slope to be used to find the normal line is

- \frac{1}{m}

- \frac{1}{m} = - \frac{1}{- 12} = \frac{1}{12}

Determine now the Normal Line using

y - y_{1} = (\frac{1}{12}) (x - x_{1})

y - (- 8) = (\frac{1}{12}) (x - 2)

after simplification the final answer:
x-12y=98 or

y = \frac{x}{12} - \frac{49}{6}

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