Question

What is the equation of the line that is normal to `f(x)= ln(x^2-x)` at `x= 1`?

Answer 1

There is no normal as `f(1)` is not defined

Explanation:

`f(x)=ln(x^2-x)`
As, `x -> 1 => f(1)->-oo`
graph{ln(x^2-x) [-10, 10, -5, 5]}

Hence, the value of f(x) has a vertical asymptote at `x=1`, so we could argue that the equation of the tangent as `x->1` is the vertical line `x=1`, but this would lead to a theoretically infinite number of normals