What is the equation of the normal line of f(x)=16x4−x3 at x=4?

Writingf(x)=(16x4−x3)12then we get by the power and chain rulef′(x)=12⋅(16x4−x3)−12(64x3−3x2)so we getf′(4)=2533⋅7and the slope of the normal line is given bymN=−37253andf(4)=247so our equation hase the formy=−3⋅7253x+nplugging x=4, y=247 in this equation to get nn=427+12⋅7253

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High School
Calculus and Analysis
Calculus 1
Derivatives

Vanessa Mccarty

Answered question

2022-04-10

What is the equation of the normal line of

f (x) = \sqrt{16 x^{4} - x^{3}}

at x=4?

Answer & Explanation

Ferrito90gn

Beginner2022-04-11Added 12 answers

Writing

f (x) = {(16 x^{4} - x^{3})}^{\frac{1}{2}}

then we get by the power and chain rule

f^{'} (x) = \frac{1}{2} \cdot {(16 x^{4} - x^{3})}^{- \frac{1}{2}} (64 x^{3} - 3 x^{2})

so we get

f^{'} (4) = \frac{253}{3 \cdot \sqrt{7}}

and the slope of the normal line is given by

m_{N} = - \frac{3 \sqrt{7}}{253}

and

f (4) = 24 \sqrt{7}

so our equation hase the form

y = - \frac{3 \cdot \sqrt{7}}{253} x + n

plugging x=4,

y = 24 \sqrt{7}

in this equation to get n

n = 42 \sqrt{7} + \frac{12 \cdot \sqrt{7}}{253}

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