Is the range correct? What is the range, using interval notation, of:
$$
f(x) = \frac{2x-4}{x^2+x-2}
$$
The answer I have is 
$$
(-\infty, 2/9] \cup [2, \infty)
$$ 
but i am not sure if it is correct. 
 A: Make up the quadratic equation for $x$:
$$y=\frac{2x-4}{x^2+x-2} \Rightarrow yx^2+(y-2)x+4-2y=0.$$
The equation will have a real solution when its discriminant is non-negative:
$$(y-2)^2-4y(4-2y)\ge 0 \Rightarrow 9y^2-20y+4\ge 0 \Rightarrow y\in (-\infty,\frac29\big{]}\cup [2,+\infty).$$
A: First, you decompose your function w.r.t. the poles 
$$
f(x)=\frac{2x-4}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}
$$
you get $A+B=2, 2A-B=-4$ hence $3A=-2, A=-\frac{2}{3}, B=\frac{8}{3}$. The derivative of $f$ is 
$$
\frac{-2 x^2 + 8 x}{x^4 + 2 x^3 - 3 x^2 - 4 x + 4}
$$
Now, as the denominator is $((x-1)(x+2))^2$ the derivative has the sign of its numerator $-2 x^2 + 8 x=-2x(x-4)$ and we get the variation table 
$$
\begin{array}{c|c|c|c|c|c|c|c|c}
x        & (-\infty,-2)& -2 &(-2, 0) & 0 & (0,1) &1  & (1,4) & 4 & (4,+\infty)
\\
sign(f') &  -          & ND &  -     & 0 &  +    & ND &  +  & 0 & -
\\
f   & 0\searrow -\infty & ND &+\infty\searrow 2 & 2 & 2\nearrow +\infty & ND & -\infty\nearrow \frac{2}{9}&\frac{2}{9} &\frac{2}{9} \searrow -\infty
\end{array}
$$
ND means ``Non Defined''.
Hope it helps   
