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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.

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    $\begingroup$ I'm sorry if I haven't explained myself very clearly. English isn't my first language. $\endgroup$ – user111913 Nov 27 '13 at 20:11
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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 $$D= \{x: \;x \in \mathbb{R}, x\neq 0 \} $$

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

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