<s>[INST] Find the greatest Integer $c$ such that $f(x)$ does not belong to $[-1 \:\: \frac{-1}{3}]$ Find the Greatest integer $c$ such that the function $$f(x)=\frac{x-1}{c-x^2+1}$$ does not take the values in the interval
 $[-1 \:\: \frac{-1}{3}]$
I have taken $$y=\frac{x-1}{c-x^2+1}$$ and by Re arranging we get
$$yx^2+x-((c+1)y+1)=0$$ from which
$$x=\frac{-1\pm \sqrt{4(c+1)y^2+4y+1}}{2y}=\frac{-1 \pm \sqrt{g(y)}}{2y}$$ where 
$$g(y)=4(c+1)y^2+4y+1$$
For any $c$ , $g(-1)=4c+1$ and $g(\frac{-1}{3})=\frac{4c+1}{9}$
But i have no clue to proceed further.
 [/INST] Let $$\displaystyle t = \frac{x-1}{c-x^2+1}\Rightarrow tx^2+x-1-t(c+1)=0$$
so $$1+4t[1+(c+1)t]\geq 0\Rightarrow 4ct^2+(2t+1)^2\geq 0$$
above inequality is true for $t=0$
now when $t\neq 0,$ then substitute $\displaystyle \frac{1}{2t} = -(y=1), y\neq \bigg[-\frac{1}{2},\frac{1}{2}\bigg]$
we have $c+y^2\geq 0$(so $c$ must be negative otherwise it is true for all $y$)
so $$y\in \bigg(-\infty,-\sqrt{-c}\bigg]\cup \bigg[\sqrt{-c},\infty\bigg)$$
so $\displaystyle -\sqrt{-c}<-\frac{1}{2}$ and $\displaystyle \sqrt{-c}>\frac{1}{2}$
at last we have $$\displaystyle -c>\frac{1}{4}\Rightarrow c<-\frac{1}{4}$$
$$\lfloor c \rfloor _{max} = -1$$
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