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

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Answer 1 · 2018-04-14T02:52:22

$x=0$

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

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

$f'(0)=0$ 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

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