# How do you find the domain and range of y = sqrt (1-x^2) / x?

Oct 21, 2015

Domain: $\left[- 1 , 0\right) \cup \left(0 , 1\right]$
Range: $\left(- \infty , + \infty\right)$

#### Explanation:

Right from the start, you know that the denominator of the fraction cannot be equal to zero. This means that the domain of the function will not include the values of $x$ that will make the denominator equal to zero.

Moreover, since you're dealing with a radical term, you need to make sure that the domain of the function does not include values of $x$ that will make the expression under the square root negative.

This means that you also need

$1 - {x}^{2} \ge 0$

${x}^{2} \le 1 \implies x \le \pm 1$

This means that the domain of the function will be $\left[- 1 , 0\right) \cup \left(0 , 1\right]$.

To get the range of the function, find its inverse function

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

Switch $y$ and $x$, then solve for $y$

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

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

Square both sides of the equation to get

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

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

Rearrange to get

${y}^{2} \cdot \left({x}^{2} + 1\right) = 1$

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

Finally, take the square root of both sides to get

$\sqrt{{y}^{2}} = \sqrt{\frac{1}{{x}^{2} + 1}}$

$y = \frac{1}{\sqrt{{x}^{2} + 1}}$

Notice that the denominator cannot be equal to zero, since ${x}^{2} \ge 0 , \forall x \in \left[- 1 , 0\right) \cup \left(0 , 1\right]$. This means that you don't have restrictions for the range of the function, and so the range will be $x \in \mathbb{R}$, or $\left(- \infty , + \infty\right)$.

graph{sqrt(1-x^2)/x [-10, 10, -5, 5]}