Question

What is the range of the function `y = sqrt(x^2+1)-x` ?

Answer 1

Use an algebraic method to find range is `(0, oo)`

Explanation:

If a particular value of `y` is in the range, then we can solve for `x` to give that value of `y`.

Suppose:

`y = sqrt(x^2+1)-x`

Adding `x` to both sides we get:

`y+x = sqrt(x^2+1)`

Squaring both sides (which may introduce spurious solutions) we get:

`y^2+2xy+x^2 = x^2+1`

Subtract `x^2+y^2` from both sides to get:

`2xy = 1-y^2`

Note that we must have `y != 0` since otherwise this equation would become `0 = 1`. So in particular `y=0` is not in the range.

Divide both sides by `2y` to get:

`x = (1-y^2)/(2y)`

So this is a requirement. To see if it is sufficient, substitute it back in the left hand side of the original equation:

`sqrt(x^2+1)-x`

`= sqrt(((1-y^2)/(2y))^2+1)-(1-y^2)/(2y)`

`= sqrt(((1-2y^2+y^4)+4y^2)/(4y^2))-(1-y^2)/(2y)`

`=sqrt((1+y^2)^2/(4y^2))-(1-y^2)/(2y)`

`=(1+y^2)/(2abs(y))-(1-y^2)/(2y)`

If `y > 0` then `abs(y) = y` and this becomes:

`(1+y^2)/(2y)-(1-y^2)/(2y) = (2y^2)/(2y) = y`

If `y < 0` then `abs(y) = -y` and this becomes:

`-(1+y^2)/(2y)-(1-y^2)/(2y) = -2/(2y) = -1/y != y`

So there is a solution for `x` if and only if `y > 0`.

So the range is `(0, oo)`