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

Aug 7, 2017

Regardless of the value of $k$, $f \left(x\right)$ will always have at least one zero.

Explanation:

Given:

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

For any given value of $k$, this function has domain $\mathbb{R} \text{\} \left\{k\right\}$. 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:

${x}^{2} - 1 = {x}^{2} - {1}^{2} = \left(x - 1\right) \left(x + 1\right)$

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

Hence we have three cases:

• $k = 1$
in which case $f \left(- 1\right) = 0$

• $k = - 1$
in which case $f \left(1\right) = 0$

• $k \ne 1$ and $k \ne - 1$
in which case both $f \left(- 1\right) = 0$ and $f \left(1\right) = 0$