How do you find the relative extrema for #f(x) = (x^2 - 3x - 4)/(x-2)#?

1 Answer
May 15, 2017

That function has no relative extrema.

Explanation:

#f(x) = (x^2 - 3x - 4)/(x-2)#

Domain of #f# is #(-oo,2)uu(2,oo)#

#f'(x) = (x^2 - 4x +10)/(x-2)^2#

#f'(x)# exists for all #x# in the domain of #f#

#f'(x) = 0# where #x^2-4x+10 = 0#. But this quadratic has discriminant #16-40# which is negative. Therefore, there are no real solutions to the equation, and consequently there are no critical numbers for #f#.

Since relative extrema occur at critical numbers, there are no relative extrema for #f#.