How do you find the domain and range of #f(x) = x/(x^2 +1)#?

2 Answers
May 18, 2018

Answer:

The domain of #f# is #RR#,
and the range is #{f(x) in RR:-1/2<=f(x)<=1/2}#.

Explanation:

Solving for the domain of #f#, we will observe that the denominator is always positive, regardless of #x#, and indeed is least when #x=0#. And because #x^2>=0#, no value of #x# can give us #x^2=-1# and we can therefore rid ourself of the fear of the denominator ever equalling naught. By this reasoning, the domain of #f# is all real numbers.

By contemplating the output of our function, we will notice that, from the right the function is decreasing until the point #x=-1#, after which the function steadily increases. From the left, it is the opposite: the function is increasing until the point #x=1#, after which the function steadily decreases.

From either direction, #f# cannot ever equal #0# except at #x=0# because for no number #x>0 or x<0# can #f(x)=0#.

Therefore the highest point on our graph is #f(x)=1/2# and the lowest point is #f(x)=-1/2#. #f# can equal all numbers in between though, so the range is given by all real numbers in between #f(x)=1/2# and #f(x)=-1/2#.

May 18, 2018

Answer:

The domain is #x in RR#. The range is #y in [-1/2, 1/2]#

Explanation:

The denominator is

#1+x^2>0, AA x in RR#

The domain is #x in RR#

To find, the range procced as follows :

Let #y=x/(x^2+1)#

#y(x^2+1)=x#

#yx^2-x+y=0#

In order for this quadratic equation to have solutions, the discriminant #Delta >=0#

Therefore,

#(-1)^2-4*y*y>=0#

#1-4y^2>=0#

The solution to this inequality is

#y in [-1/2, 1/2]#

The range is #y in [-1/2, 1/2]#

graph{x/(x^2+1) [-3, 3.93, -1.47, 1.992]}