Question

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

Answer 1

`x=0`

Explanation:

`f(x)=4/(x^3-1)=4(x^3-1)^(-1)`
`f'(x)=-4(x^3-1)^(-2)(3x^2)`

`f'(0)=-4(0^3-1)^(-2)(3(0)^2)=0`

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

since the point `(0,f(0))` lies on the vertical line `x=0`, that is the normal line