Question

What is the domain and range of `f(x) = 1/(1+x^2)`?

Answer 1

Domain: `(-oo, +oo)`
Range: `(0,1]`

Explanation:

The only possible restriction for this function's domain is when the denominator is equal to zero, but since `x^2>0, (AA)x` in `(-oo, +oo)`, `1 + x^2!` will always be positive.

This means that the domain of the function will be `RR`, or `(-oo, +oo)`.

Since the denominator will always be positive, `1/(1+x^2` will always be positive as well.

The range of the function will be `y>0` and `y<=1`, since the maximum value of `f(x)` occurs at `x=0`.

`f(0) = 1/(1 + 0) = 1 ->` global maximum

That happens because you have `f(x)<1, (AA)x !=0`.

The range of the function will thus be `(0, 1]`.

graph{1/(1+x^2) [-8.89, 8.89, -4.444, 4.445]}