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

May 9, 2016

Use an algebraic method to find range is $\left(0 , \infty\right)$

#### 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:

${y}^{2} + 2 x y + {x}^{2} = {x}^{2} + 1$

Subtract ${x}^{2} + {y}^{2}$ from both sides to get:

$2 x y = 1 - {y}^{2}$

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

Divide both sides by $2 y$ to get:

$x = \frac{1 - {y}^{2}}{2 y}$

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

$\sqrt{{x}^{2} + 1} - x$

$= \sqrt{{\left(\frac{1 - {y}^{2}}{2 y}\right)}^{2} + 1} - \frac{1 - {y}^{2}}{2 y}$

$= \sqrt{\frac{\left(1 - 2 {y}^{2} + {y}^{4}\right) + 4 {y}^{2}}{4 {y}^{2}}} - \frac{1 - {y}^{2}}{2 y}$

$= \sqrt{{\left(1 + {y}^{2}\right)}^{2} / \left(4 {y}^{2}\right)} - \frac{1 - {y}^{2}}{2 y}$

$= \frac{1 + {y}^{2}}{2 \left\mid y \right\mid} - \frac{1 - {y}^{2}}{2 y}$

If $y > 0$ then $\left\mid y \right\mid = y$ and this becomes:

$\frac{1 + {y}^{2}}{2 y} - \frac{1 - {y}^{2}}{2 y} = \frac{2 {y}^{2}}{2 y} = y$

If $y < 0$ then $\left\mid y \right\mid = - y$ and this becomes:

$- \frac{1 + {y}^{2}}{2 y} - \frac{1 - {y}^{2}}{2 y} = - \frac{2}{2 y} = - \frac{1}{y} \ne y$

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

So the range is $\left(0 , \infty\right)$