<s>[INST] Prove that $f(x)=\frac{1+11x-6x^2}{(x-3)(1-2x)}$ is a decreasing function How do I prove that $f(x)$ is a decrasing function
$$f(x)=\frac{1+11x-6x^2}{(x-3)(1-2x)},x>3$$

I did:
$$f{'(x)}=-\frac{4}{(x-3)^2}+\frac{2}{(1-2x)^2}$$
$$f^{'}(x)<0$$
then I got
$$-\frac{7}{9}<x<1$$
So where do I go from here
 [/INST] \begin{align}f(x) &= \frac{1+11x-6x^2}{(x-3)(1-2x)}\\&=\frac{3(-2x^2+7x-3)+11x-21x+1+9}{(x-3)(1-2x)}\\
&=3 + \frac{10(1-x)}{(x-3)(1-2x)}\\
&=3 + \frac{4}{x-3}- \frac{2}{1-2x}\\
&= 3 + \frac4{x-3}+\frac{1}{x-0.5}\end{align}
For $x>3$, as $x$ increases, we can see that $f(x)$ decreases.
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