Question

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

Answer 1

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`.