Question

For what values of `k` will the function `f(x) = (x^2-1)/(x-k)^2` have a zero?

Answer 1

Regardless of the value of `k`, `f(x)` will always have at least one zero.

Explanation:

Given:

`f(x) = (x^2-1)/(x-k)^2`

For any given value of `k`, this function has domain `RR "\" { k }`. That is: It is well defined for all values of `x` except `k`, where it has a vertical asymptote.

The numerator can be factored as a difference of squares:

`x^2-1 = x^2-1^2 = (x-1)(x+1)`

This takes the value `0` if `x=1` or `x=-1`.

Hence we have three cases:

• `k = 1`
  in which case `f(-1) = 0`

• `k = -1`
  in which case `f(1) = 0`

• `k != 1` and `k != -1`
  in which case both `f(-1) = 0` and `f(1) = 0`