Question

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

Answer 1

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

Explanation:

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`