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`.