# What is the domain of #f(x)=(x+3)/sqrt(x^2-9) #?

##### 1 Answer

#### Answer:

Domain:

#### Explanation:

The domain of the function will include any value of **negative**.

For real numbers, you can only take the square root of positive numbers, which means that

#x^2 - 9 >=0#

SInce you also need this expression to be different from zero, you get

#x^2 - 9 > 0#

#x^2 - 3^2 > 0#

#(x-3)(x+3) > 0#

This inequality is true when you have both terms *negative* or both terms *positive*. For values of

#{(x-3 < 0), (x + 3<0) :} implies (x-3)(x+3) > 0#

For values of

#{(x-3>0), (x + 3 > 0) :} implies (x-3)(x+3) > 0#

This means that **any** value of **smaller** than **greater** than **not** satisfy this inequality.

This means that the domain of the function will be