# How would you find the domain of each: #y=(3x+2)/(4x+1)#? #y=(x^2-4)/(2x+4)#? #y=(x^2-5x-6)/(x^2+3x+18)#? #y=(2^(2-x))/(x)#? #y=sqrt(x-3)-sqrt(x+3)#? #y=sqrt(2x-9)/(2x+9)#?

##### 2 Answers

#### Answer:

The domain would be all real numbers except for any value of

#### Explanation:

For example, we can look at your first statement,

Any value of

To find what that value is, you can set up an equation like this:

After solving this, you would get the answer of

You would do the same for the other fraction problems. In the second problem, you would need to factor the denominator or use the quadratic formula to find the non-domain values.

For your square root problems, the domain would be any value of

For example, let's look at

We can set up our equations again:

The answers here would be

This means that any value of

The last question is an amalgam of these two problem types: *or* anything that will make the expression in the square root negative.

We can set up equations here again:

So anything less than

I hope that makes sense and helps you solve the fraction problems that I did not address.

#### Answer:

Please see below.

#### Explanation:

Domain in

For example if we have

In case we have a quadratic polynomial such as

Sometimes factors may cancel out, if they are common between numerator and denominator. In that case, we call it a hole, because though

- In
#y=(3x+2)/(4x+1)# , domain isall#x# other than#x=-1/4# , as latter makes denominator#0# . - In
#y=(x^2-4)/(2x+4)=((x+2)(x-2))/(2(x+2))=(x-2)/2# and we cannot have#x=+-2# and domain is values of#x# other than#+-2# and at#x=2# , we have a hole. - As
#y=(x^2-5x-6)/(x^2+3x+18)=((x-6)(x+1))/((x+3/2)^2+63/4)# . Note that least value of denominator is#63/4# and hence#y=f(x)# exists for all vales of#x# and hence domain is#RR# . - In
#y=2^(2-x)/x# , we have no restrictions on#x# as far as numerator is concerned, however, denominator restricts domain of#x# so that#x!=0# . - In
#y=sqrt(x-3)-sqrt(x+3)# , as we have#sqrt(x-3)# we cannot have#x<3# and as we also have#sqrt(x+3)# , other restriction is we cannot have#x<-3# . This means domain is#x>=3# . - As
#y=sqrt(2x-9)/(2x+9)# , numerator places a restriction that#x>=9/2# and denominator#x!=-9/2# . This can be combined and domain is#x>=9/2# .