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

Jun 24, 2018

$y = - \frac{3 \cdot \sqrt{7}}{253} x + \frac{12 \cdot \sqrt{7}}{253} + 24 \cdot \sqrt{7}$

Explanation:

Writing

$f \left(x\right) = {\left(16 {x}^{4} - {x}^{3}\right)}^{\frac{1}{2}}$
then we get by the power and chain rule

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

so we get

$f ' \left(4\right) = \frac{253}{3 \cdot \sqrt{7}}$
and the slope of the normal line is given by

${m}_{N} = - \frac{3 \sqrt{7}}{253}$
and

$f \left(4\right) = 24 \sqrt{7}$

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 $n$

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