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

1 Answer
Aug 7, 2017

Answer:

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#