Let #y=f(x) = (x^2 - 5x)/5#. Then #f# is differentiable, since we can break it up like #1/5x^5 - x#, which is a polynomial, and infinitely differentiable. Its first derivative is:

#d/dx(1/5x^5 - x) = x^4 - 1#. We want to find the roots of this expression, or where the derivative is zero:

#x^4-1 = 0 => (x^2 - 1)(x^2 + 1) = 0 => (x-1)(x+1)(x^2+1) = 0#

We can see that for #x = -1# or #x = 1#, #dy/dx = 0#.

#x^4 - 1# is a polynomial with two roots. Since the coefficient of #x^4# is #1 > 0#, it holds that:

#x^4 - 1# is positive, when #x in (-infty, -1)uu(1, +infty)#

#x^4 - 1# is negative, when #x in (-1, 1)#

When #dy/dx > 0#, #f# is increasing.

When #dy/dx < 0#, #f# is decreasing.

Therefore,

#f# is increasing, when #x in (-infty, -1)uu(1, +infty)#

#f# is decreasing, when #x in (-1, 1)#

(Since #f#'s domain is #RR#, we can safely extend our analysis to the entirety of #(-infty, +infty)#).