# What is the equation of the line normal to f(x)=4/(x^3-1)  at x=0?

Apr 14, 2018

$x = 0$

#### Explanation:

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

$f ' \left(0\right) = - 4 {\left({0}^{3} - 1\right)}^{- 2} \left(3 {\left(0\right)}^{2}\right) = 0$

$f ' \left(0\right) = 0$ means the tangent line is horizontal
the normal line then must be vertical (line is in the form $x = a$)

since the point $\left(0 , f \left(0\right)\right)$ lies on the vertical line $x = 0$, that is the normal line