I am trying to solve for the exhaustive set of values of $p$ so that the interval $[-1,-1/3]$ does not fall into the range of the function $f(x)$ given by the following expression with $x\ne\pm\sqrt{p+1}$.
$$f(x)=\frac{x-1}{p-x^2+1}$$
Now usually the procedure involved in solving this type of problems is making a quadratic in $x$, in terms of $y$ and setting some condition on the discriminant.
$$yx^2+x-(y(p+1)+1)=0\\ D=1+4y(y(p+1)+1)$$
It is not clear as to what the condition should be imposed on the discriminant. Any hints are appreciated. Thanks.
Notice that $f(x) \to 0^-$ as $x\to \infty$, and $f(x) \to 0^+$ as $x\to -\infty$. Also, it is continuous for $p\leq 0$ so the only way that your condition will be true is if the global minimum value is greater than $-\frac 13$.
$$f’(x) = \frac{(p-x^2-1) - (x-1)(-2x)}{(p-x-1)^2} =0 \implies x = 1 \pm \sqrt{-p}$$ Note that if $p\gt 0$ then there would be no local maxima/minima suggesting the existence of a vertical asymptote, and hence our condition would not be satisfied. So, $p\leq 0$. Now it is easy to check that $f(x)$ attains it’s minimum value at $x=1+\sqrt{-p}$ and we want
$$ f(1+\sqrt{-p}) \gt -\frac 13 $$ $$\frac{\sqrt{-p}}{2p - 2\sqrt{-p}} \gt -\frac 13$$ $$3\sqrt{-p} \lt 2\sqrt{-p} -2p$$ $$-p \lt 4p^2$$ $$\implies \color{blue}{p\lt -\frac 14}$$ and we are done.