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

##### 1 Answer

#### Answer:

Domain:

Range:

#### 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

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 **negative**.

This means that you also need

#1 - x^2 >= 0#

#x^2 <= 1 implies x <= +- 1#

This means that the domain of the function will be

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

#y = sqrt(1-x^2)/x#

Switch

#x = sqrt(1-y^2)/y#

#xy = sqrt(1 - y^2)#

Square both sides of the equation to get

#(xy)^2 = (sqrt(1-y^2))^2#

#x^2y^2 = 1 - y^2#

Rearrange to get

#y^2 * (x^2 + 1) = 1#

#y^2 = 1/(x^2 + 1)#

Finally, take the square root of both sides to get

#sqrt(y^2) = sqrt(1/(x^2 + 1))#

#y = 1/sqrt(x^2 + 1)#

Notice that the denominator **cannot** be equal to zero, since

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