The easiest way I Know is to equate the given function[f(x)] to “y”.
For Example......
x/(x2 + 1 ) = y
i.e. x = y(x2 ) + y
i.e. y(x2 ) – x + y = 0
This is a Quadratic Equation in terms of “x”
And as it is understood that “x” is real.... The determinant of this equation must be

O.....
Therefore (-1)
2 – 4(y)(y)

0
1 – 4y
2 
0
Thus 4y
2 – 1

0
y
2 
(1/4)
Therefore.... y

(-
∞ , -1/2)
(1/2 , ∞)This gives the Range of the given function........ i.e. (-
∞ , -1/2)
(1/2 , ∞)