Question 937605
f(x)=x+1/x  is ambiguous.  Further it is not a quadratic function.  You can get better help asking for what you exactly want and giving the full problem description.


Some clarification given:
like {{{y=x+1/x}}}, which was exactly as you wrote.  Sometimes students want the x+1 grouped as a separate expression but omit the grouping symbols.


The allowed values for x must be {{{x<>0}}}, so that y will be undefined for x at 0.


You wanted to find the range.  Think what happens to y very near to 0 on the left and on the right.  ....
....  UNBOUNDED negative as x approaches 0 from the left, and unbounded positive as x approaches 0 from the right.


Without further algebra, although you might want it,...  I skip ahead to a graph.


{{{graph(300,300,-15,15,-15,15,x+1/x)}}}


A derivative might be needed to find the rest of the range restriction.
THIS MEANS YOU ARE  OR HAVE STUDIED CALCULUS 1.  ( I cannot think of another way).


{{{d/dy=d(x+1/x)}}}
{{{1+(-1)x^(-2)}}}
{{{1-1/x^2}}}
{{{1(x^2/x^2)-1/x^2}}}
{{{(x^2-1)/x^2}}}


WHERE is this derivative equal to 0?
{{{x^2-1=0}}}
{{{x^2=1}}}
{{{highlight_green(x=-1)}}}  or  {{{highlight_green(x=1)}}}


Note from the sketch of the graph you might make,  x=-1 will give a local max, and x=1 will give a local min.  Use your given function definition to find those function values.  The range must EXCLUDE the values between f(-1) and f(1).


To finish....
Range is the set of {{{f(x)<=-2}}}  and {{{f(x)>=2}}}.