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

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Answer 1 · 2015-11-03T23:35:59

The domain of $f(x)$ is the whole of $RR$ , that is $(-oo, oo)$ in interval notation.

The range of $f(x)$ is $(0, 1]$ .

$1+x^2 >= 1 > 0$ for all $x in RR$ .

So the denominator is never $0$ and $f(x)$ is well defined for all $x in RR$ . So the domain of $f(x)$ is $RR = (-oo, oo)$

Let $y = f(x) = 1/(1+x^2)$

Multiplying both sides by $(1+x^2)$ we get:

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

This has no solution if $y = 0$ , so $0$ is not in the range of $f(x)$ .

If $y != 0$ then we can divide both sides by $y$ to get:

$1+x^2 = 1/y$

So: $x^2 = 1/y-1$

Since $x^2 >= 0$ for all $x in RR$ , we require $1/y-1 >= 0$

If $y > 1$ then $0 < 1/y < 1$ , so $1/y - 1 < 0$ . Therefore $(1, oo)$ is not part of the range.

If $y < 0$ then $1/y < 0$ and $1/y - 1 < 0$ . So $(-oo, 0)$ is not part of the range either.

Suppose $y > 0$ and $y <= 1$

Since $y > 0$ we can divide both sides of the second inequality by $y$ to get:

$1 <= 1/y$

Then subtract $1$ from both sides to get:

$1/y-1 >= 0$ as required.

Hence $x = +-sqrt(1/y-1)$ are two values that give $f(x) = y$ .

So the range includes the whole of $(0, 1]$

graph{1/(1+x^2) [-5, 5, -2.5, 2.5]}

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