Range of function doesn't contain any value belonging to an interval

Solution 1:

Ultimately, $y$ being in the range means that a certain quadratic equation has real solutions. This quadratic equation is $x-1=(-x^2+1+p)y$. You can rearrange this to be of the form $p_y(x)=0$. So the range of the rational function is the solution to the nonlinear inequality $\Delta(y) \geq 0$ where $\Delta(y)$ is the discriminant of $p_y$. The result will be an inequality of the form $q(y) \geq 0$ where $q$ is another quadratic polynomial. These are pretty easy to solve, but there is some case work:

1. If the coefficient of $y^2$ is positive, then the solution to $q(y) \geq 0$ is $(-\infty,r_1] \cup [r_2,\infty)$, where $r_1<r_2$ are the roots of $q$. If $q$ has no roots, then the solution is just $\mathbb{R}$.
2. If the coefficient of $y^2$ is negative, then the solution to $q(y) \geq 0$ is $[r_1,r_2]$ under the same convention. If $q$ has no roots, then there are no solutions at all.

It will turn out here that you are in case 1.

Solution 2:

You want $4y^2(p+1)+4y +1 \ge 0$ to have a solution $y\in (-\infty,-1)\cup (-\frac13,+\infty)$ with roots $y_1=-1,y_2=-\frac13$. Hence: $$y_1y_2=\frac{1}{4(p+1)}=\frac13 \Rightarrow p=-\frac14.$$ For the range stated above the values of $p$ must be $p<-\frac14$.

Solution 3:

Let $$\displaystyle t = \frac{x-1}{p-x^2+1}\Rightarrow tx^2+x-1-t(p+1)=0$$

so $1+4t[1+(p+1)t]\geq 0$

$\Rightarrow 4pt^2+(2t+1)^2\geq 0$

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 $p+y^2\geq 0$(so $p$ must be negative otherwise it is true for all $y$)

so $$y\in \bigg(-\infty,-\sqrt{-p}\bigg]\cup \bigg[\sqrt{-p},\infty\bigg)$$

so $\displaystyle -\sqrt{-p}<-\frac{1}{2}$ and $\displaystyle \sqrt{-p}>\frac{1}{2}$

at last we have $$\displaystyle -p>\frac{1}{4}\Rightarrow p<-\frac{1}{4}$$