Range of x + 1/x ?

Question

8983 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-22T12:28:31

#(-oo,-2]uu[2,oo)#

The range of a function is defined as all possible values of #f(x)# or #y#. It can also be defined as the domain of #f^-1(x)#.

To find #f^-1(x)#:

#y=x+1/x#

Switch the variables, and then solve for #y#.

#x=y+1/y#

#xy=y^2+1#

#y^2-xy+1=0#

The above is a quadratic equation of the form #ay^2+by+c=0#

For such quadratic equations, #y=(-b+-sqrt(b^2-4ac))/(2a)#

Here, #a=1, b=-x, c=1#. Inputting:

#y=(-(-x)+-sqrt((-x)^2-4*1*1))/(2*1)#

#y=(x+-sqrt(x^2-4))/2#

#y=(x+sqrt(x^2-4))/2, (x-sqrt(x^2-4))/2=f^-1(x)#

We must find the domain of the above. When the denominator, a radical, or a logarithmic function is less than zero, the function is undefined.

So here, for both values (as only the radical, the same in both values, matters) we have:

#sqrt(x^2-4)>=0#

#x^2-4>=0#

#x^2>=4#

#x>=2# and #x<=-2#

So #x# is undefined between #-2# and #2#.

Therefore, the domain of #f^-1(x)# is #(-oo,-2]uu[2,oo)#

By extension, the range of #f(x)# is #(-oo,-2]uu[2,oo)#.