Find the largest possible domain of this rational function!

Find the largest possible domain of the function
f(x) = 1/sqrt(4+3x)
Express your answer in interval notation.

Thanks!

4 + 3x must be > 0
3x > - 4
x > -4/3

So domain is ( -4/3 , infinity)

As an aside, "largest possible domain" is strange terminology to me; I've always known the domain of a function to be all possible values of x, so prefixing it with "largest possible" sounds redundant to me.

But I digress.

f(x) = 1/sqrt(4 + 3x)

First off, this function has two restrictions.
1) The function has a radical (square root) in it, and
2) The function is a rational (i.e. a fraction).

Restrictions of a radical: Inside of the square root must be greater than or equal to 0, since we cannot take the square root of a negative number to get a real result.
Restrictions of a rational: Denominator cannot equal 0.

In order for this function to be valid, the inside of the radical must be greater than equal to 0, i.e.
4 + 3x >= 0

If we solve this inequality, we get
3x >= -4
x >= -4/3

Next, in order for this function to be valid, the denominator cannot equal 0. So what then makes the denominator _equal_ to 0? If we find this out, we'll know what values x CANNOT be.

sqrt(4 + 3x) = 0. Square both sides to get
4 + 3x = 0
3x = -4
x = -4/3; therefore, x CANNOT equal -4/3.

Combining the two facts:

x >= -4/3 AND x cannot equal -4/3.
Which is the same as saying

x > -4/3

Domain: (-infinity, -4/3)

Or, in set notation,
{x | x > -4/3, for x E R}

"The set of all real numbers x such that x is strictly greater than -4/3."

4 + 3x >= 0
3x >= -4
x >= -4/3

So, x is greater than or equal to -4/3 or you could say (-4/3, infinity) in interval notation.

sqrt(4+3x)>0
x>-4/3 or x belongs (-4/3, +infinity)