# Find the domain of the function $f(x)= 3x^2+4/2x-1$ (If they exist)

I'm having trouble understanding how to find the domain of the above function.

I've done $2x-1=0$ and got $x=1/2$ but I'm not entirely sure where to take it from there.

Is that right? any advice greatly appreciated.

• I'm sorry if I haven't explained myself very clearly. English isn't my first language. – user111913 Nov 27 '13 at 20:11

It is right if you mean $$f(x)=\frac {3x^2+4}{(2x-1)} \:or \:f(x)=3x^2+\frac {4}{(2x-1)}$$Then domain is $$D= \{x:\;x \in \mathbb{R}, x\neq \frac12 \}$$ But if $$f(x)=3x^2+\frac {4}{(2x)}-1$$ then the domain is $$D= \{x: \;x \in \mathbb{R}, x\neq 0 \}$$

• Yeah its the first function you mentioned. Wasn't sure how to explain the domain. thank you for your assistance. greatly appreciated – user111913 Nov 27 '13 at 20:41

Since at $x = \frac{1}{2}$ the function is undefined, it's typical to say that the domain is "all real numbers except $\frac{1}{2}$" or, more formally,

$$\{ x : x \in {\mathbb R} \text{ and } x \ne \frac{1}{2} \}.$$

Does that help?