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)#?

For example, we can look at your first statement, #(3x+2)/(4x+2)#.

Any value of #x# that makes the denominator equal to 0 cannot be in the domain because you can't divide by zero.
To find what that value is, you can set up an equation like this:
#4x + 2 = 0#

After solving this, you would get the answer of #x = -1/2#, which means that #x# cannot equal #-1/2#, or the denominator will equal 0. So the domain for this one would be written #x: x in RR, x != -1/2#

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 #x# except for the ones that make the expression inside the #sqrt# negative, since you can't take the square root of a negative number.

For example, let's look at #y = sqrt(x - 3) - sqrt(x + 3)#
We can set up our equations again:
#x - 3 = 0#
#x + 3 = 0#

The answers here would be #x = 3# and #x = -3#.
This means that any value of #x# that is lower than 3 (this would include -3, so we don't need to make any special considerations) will leave you with a value less that 0. None of these values can be part of the domain.
#x: x in RR, x >= 3#

The last question is an amalgam of these two problem types: #x# can not equal anything that will make the denominator equal 0 or anything that will make the expression in the square root negative.

We can set up equations here again:
#2x + 9 = 0# --> #x = -9/2#
#2x - 9 = 0# --> #x = 9/2#

So anything less than #9/2# will make the square root negative, and #-9/2# will make the denominator 0. Since #-9/2 < 9/2#, we don't need to add in an extra statement for it:
#x: x in RR, x >= 9/2#

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

Domain in #y=f(x)# means thee values which #x# can take. For finding domain we normally start in a reverseway i.e. values which #x# cannot take.

For example if we have #x-a# in denominator, as we cannot have denominator as #0#, we cannot have #x=a#. Similarly, if we have #sqrt(x-a)# as we cannot have square root og a negative number, we cannot have #x-a<0# or #x < a#.

In case we have a quadratic polynomial such as #ax^2+bx+c# in denominator, we should either factorise it to #a(x-alpha)(x-beta)# or convert it to form #a(x-h)^2+k#, to check restrictions on the value of #x#.

Sometimes factors may cancel out, if they are common between numerator and denominator. In that case, we call it a hole, because though #f(x)# may be defined.

1. In #y=(3x+2)/(4x+1)#, domain isall #x# other than #x=-1/4#, as latter makes denominator #0#.
2. 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.
3. 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#.
4. 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#.
5. 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#.
6. 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#.