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

2 Answers
May 18, 2018

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

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]}