Finding domain and range I just started learning engineering functions and I'm now at Domain and range ..

I was trying this question - 

$$\frac{x+\sqrt{x+1}}{2x-1} $$

I found the domain to be 


$x$ less than $1/2$
$x$ more than $1/2$ 
$x$ more than or equal to $-1$ 


Now about finding the range , 

I'm having huge difficulty... 

I thought spitting the fractions will help but it didn't . -

Is there any way to see how to get the range easily ? Thanks
Let $t=\sqrt{x+1}\geq 0$ so $x=t^2-1$. Thus we have to find all $b\in \mathbb{R}$ that satisfy equation:

$$\frac{t^2 +t-1}{2t^2 -3} =b$$ 

If you draw a graph of rational function $f(t) = \frac{t^2 +t-1}{2t^2 -3}$, for $t\geq 0$ you will see that the range is $(-\infty,{1\over 3}]\cup ({1\over 2},\infty)$.
The domain is $[-1,+\infty)\setminus\{\frac{1}{2}\}$.

$f$ is continuous function on $\left(\frac{1}{2},+\infty\right)$, $\lim\limits_{x\rightarrow\frac{1}{2}^+}f(x)=+\infty,$ $\lim\limits_{x\rightarrow+\infty}f(x)=\frac{1}{2}$ 

and easy to see that the equation $f(x)=\frac{1}{2}$ has no solutions.

Indeed, $$f(x)=\frac{1}{2}$$ it's
$$\frac{x+\sqrt{x+1}}{2x-1}=\frac{1}{2}$$ or
$$2x+2\sqrt{x+1}=2x-1$$ or
$$2\sqrt{x+1}=-1,$$ which is impossible. 

Thus, $$f:\left(\frac{1}{2},+\infty\right)\rightarrow\left(\frac{1}{2},+\infty\right).$$

The case $x\in\left[-1,\frac{1}{2}\right)$ is the similar:

$f$ is a continuous function on $\left[-1,\frac{1}{2}\right)$, $\lim\limits_{x\rightarrow\frac{1}{2}^-}f(x)=-\infty$  and $f(-1)=\frac{1}{3}.$

Thus, $$f:\left[-1,+\frac{1}{2}\right)\rightarrow\left(-\infty,\frac{1}{3}\right).$$
Id est, the range is:
$$\left(-\infty,\frac{1}{3}\right]\cup \left(\frac{1}{2},\infty\right).$$
Done!