The function is defined for all real #x#, so let's find the extrema by differentiating and the setting the derivative equal to 0.
Differentiate:
#dy/dx = (d(1))/dx+(d(x/(x^2+1)))/dx#
The derivative of a constant is 0:
#dy/dx = 0+(d(x/(x^2+1)))/dx#
#dy/dx = (d(x/(x^2+1)))/dx#
Using the quotient rule:
#dy/dx = ((d(x))/dx(x^2+1)-x(d(x^2+1))/dx)/(x^2+1)^2#
#dy/dx = (x^2+1-x(2x))/(x^2+1)^2#
#dy/dx = (x^2+1-2x^2)/(x^2+1)^2#
#dy/dx = (1-x^2)/(x^2+1)^2#
Now that we have the first derivative, we shall set it equal to 0 and solve for the value(s) of x:
#0 = (1-x^2)/(x^2+1)^2#
Since #(x^2+1)^2≠0# for all #x#,
#0 = 1-x^2#
#x^2=1#
#x = +-1#
We can find the range by evaluating the function at #x = -1# and #x = 1#:
#y = 1 + (-1)/((-1)^2+1)# and #y = 1 + (1)/(1^2+1)#
#y = 1/2# and #y = 3/2#
The range is:
#1/2<= y <= 3/2#
We can verify this by graphing the equation:
graph{y = 1 + (x/(x^2+1)) [-10, 10, -5, 5]}