What is the equation of the normal line of $f(x)=sqrt(16x^4-x^3$ at $x=4$ ?

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Answer 1 · 2018-06-24T13:13:04

$y=-(3*sqrt(7))/253x+(12*sqrt(7))/253+24*sqrt(7)$

Writing

$f(x)=(16x^4-x^3)^(1/2)$
then we get by the power and chain rule

$f'(x)=1/2*(16x^4-x^3)^(-1/2)(64x^3-3x^2)$

so we get

$f'(4)=253/(3*sqrt(7))$
and the slope of the normal line is given by

$m_N=-(3sqrt(7))/253$
and

$f(4)=24sqrt(7)$

so our equation hase the form

$y=-(3*sqrt(7))/253x+n$

plugging $x=4,y=24sqrt(7)$ in this equation to get $n$

$n=42sqrt(7)+(12*sqrt(7))/253$

What is the equation of the normal line of f(x)=sqrt(16x^4-x^3f(x)=sqrt(16x^4-x^3 at x=4x=4?