f(x)=sqrt(x+4)

Recall that you cannot take the square root of negative numbers.

Therefore, whatever is under the square root symbol must be greater than or equal to zero.

x + 4 >= 0

Subtract 4 from each side.

x >= -4

The domain is all real numbers greater than or equal to -4.

f(x)=2x+1/x-3

I’m assuming that the whole quantity (x-3) is in the denominator. It should be written like this.

f(x)=2x+1/(x-3)

You cannot divide by zero, so x – 3 cannot equal zero. Find what value makes x-3 equal to zero, and exclude this from the domain.

x – 3 = 0

Add 3 to each side

x = 3.

The domain is all real numbers except -3.

Edit: I looked at your question in the math section.

f(x)=3x^2+5x-3

The domain of a polynomial is always all real numbers. (Polynomials don’t involve any variables in the denominator or underneath a root symbol).

RE:
Find the domain of this equation?
I don't quite get this.

f(x)=sqrt(x+4)

also one that is divided

f(x)=2x+1/x-3

I have a whole page of them but if I have a couple of examples I can figure it out from there.

first one

make f(x) to 0

square both sides

0 = (x+4)

x = -4

second one

make f(x) to 0

multiply all terms by x

0 = 2x^2 + 1 + 3x

2x^2 + 3x + 1

(2x +1)(x+1)=0

x = -1/2, x = -1