I am comfortable when I am asked to calculate the range of a rational function, but how do we do the reverse? I came across this problem.
If $$f(x)= \frac{x^2 + x + c}{x^2 + 2x + c}$$ then find the value of $c$ for which the range of $f(x)$ does not contain $[-1, -\frac{1}{3}]$.
The inequality that you got should have been $$g(y,c)=4(1-c)y^2+2(4c-2)y+(1-4c)\ge0$$Well, we want this to fail when $y$ is in this range. i.e. for $y\in[-1,-1/3]$, $$4(1-c)y^2+2(4c-2)y+(1-4c)<0$$ This means if we plot this with $y$ on the $x$-axis and the value of $g(y)$ on the $y$-axis, then it will be a positive quadratic which dips below the $x$-axis for the range of $x$ values $[-1,-1/3]$. As we vary $c$, the places where this graph cuts the $x$ axis move (i.e. the roots of $g$). Note, when $c=0$, the graph is $$g(y,c=0)=4y^2-4y+1=(2y-1)^2$$which has one repeated root at $y=1/2$. Then as $c$ is increased, the minimum point of this quadratic moves leftward. The point at which $-1/3$ joins the range of values for which $g(-1/3,c)\le0$ is when it first becomes a root. For what value of $c$ does this happen?
$$g(-1/3,c)=0\implies\cdots\implies c=\frac{25}{64}$$As we increase $c$ further, eventually $-1$ also joins this range. $$g(-1,c)=0\implies\cdots\implies c=\frac9{16}$$
For $c>\frac9{16}$, $[-1,-1/3]$ is still not in the range of $f(x)$ - but $c=\frac9{16}$ is the first value of $c$ for which this happens.
So the range of $c$ for which no points from the interval $[-1,-1/3]$ are included in the range of $f$ is given by $$\bbox[5px,border:2px solid red]{c\ge\frac9{16}}$$
Meanwhile, the range of $c$ for which at least one point from the interval $[-1,-1/3]$ is missing from the range of $f$ is given by $$\bbox[5px,border:2px solid red]{c\ge\frac{25}{64}}$$
We should probably proceed by cases, as $f$ will have a vertical asymptote and a hole for $c=0,$ a single vertical asymptote and no holes for $c=1,$ neither vertical asymptotes nor holes when $c>1,$ and otherwise has two vertical asymptotes and no holes. In all cases, $f$ has $y=1$ as its horizontal asymptote, meaning that $\lim_{|x|\to\infty}f(x)=1.$
In the $c>1$ case, differentiating yields $$f'(x)=\frac{x^2-c}{(x^2+2x+c)^2},$$ which is negative in the interval $\left(-\sqrt{c},\sqrt{c}\right),$ zero at $x=\pm\sqrt{c},$ and positive otherwise. Consequently, $f$ achieves a global maximum at $x=-\sqrt{c}$ and a global minimum at $x=\sqrt{c}.$ The minimum is the one we care about, though, since we'll require it to be greater than $-1.$ So, let's consider $$-1<f\left(\sqrt c\right)=\frac{c+\sqrt c+c}{c+2\sqrt c+c}=\frac{2c+\sqrt c}{2c+2\sqrt c}=\frac{2\sqrt c+1}{2\sqrt c+2}.$$ But this is trivially true, since $c>1$ implies that both $2\sqrt c+1$ and $2\sqrt c+2$ are positive, so that $f(c)>0>-1,$ as desired.
I leave the $c=0$ and $c=1$ cases to you.
If $c<1$ and $c\ne 0,$ we know that $x^2+2x+c=(x-j)(x-k)$ for some distinct real numbers $j$ and $k.$ We also know that $x^2+x+c$ takes on a non-zero value at each of $j,k.$ Without loss of generality, suppose that $j<k.$ Now, at $x=j,$ the denominator changes signs, while the numerator does not, meaning that we have either $$\lim_{x\nearrow j}f(x)=-\infty\text{ and }\lim_{x\searrow j}f(x)=+\infty\tag{1}$$ or $$\lim_{x\nearrow j}f(x)=+\infty\text{ and }\lim_{x\searrow j}f(x)=-\infty.\tag{2}$$ Likewise, we have either $$\lim_{x\nearrow k}f(x)=-\infty\text{ and }\lim_{x\searrow k}f(x)=+\infty\tag{3}$$ or $$\lim_{x\nearrow k}f(x)=+\infty\text{ and }\lim_{x\searrow k}f(x)=-\infty.\tag{4}$$ If $(1)$ holds, then since $f$ is continuous on $(-\infty,j)$ and since $\lim_{x\to-\infty}f(x)=1$ and $\lim_{x\nearrow j}f(x)=-\infty,$ then $(-\infty,1)$ is a subset of the range of $f,$ and so $\left[-1,-\frac13\right]$ is, too. If $(4)$ holds, then by continuity of $f$ on $(k,\infty),$ we similarly have that $\left[-1,-\frac13\right]$ is a subset of the range of $f.$ However, if $(2)$ and $(3)$ both hold, it isn't clear from what we've discussed so far whether $\left[-1,-\frac13\right]$ is a subset of the range of $f$ or not, so let's consider further when these different possibilities may occur.
The only way for $(1)$ to hold is if the numerator is negative at $x=j,$ meaning that $j$ must lie between the two zeroes of $x^2+x+c.$ In other words, by the quadratic formula, we have the following (equivalent) inequalities: $$\frac{-1-\sqrt{1^2-4c}}2<\frac{-2-\sqrt{2^2-4c}}2<\frac{-1+\sqrt{1^2-4c}}2\\-1-\sqrt{1-4c}<-2-\sqrt{4-4c}<-1+\sqrt{1-4c}\\-\sqrt{1-4c}<-1-\sqrt{4-4c}<\sqrt{1-4c}\\\left(-1-\sqrt{4-4c}\right)^2<1-4c\\1+2\sqrt{4-4c}+4-4c<1-4c\\2\sqrt{4-4c}+4<0.$$ Since this is impossible, then $(1)$ cannot hold, and so $(2)$ must hold.
Similarly, $(4)$ only holds if the numerator is negative at $x=k,$ which happens if and only if $$\frac{-1-\sqrt{1^2-4c}}2<\frac{-2+\sqrt{2^2-4c}}2<\frac{-1+\sqrt{1^2-4c}}2\\-1-\sqrt{1-4c}<-2+\sqrt{4-4c}<-1+\sqrt{1-4c}\\-\sqrt{1-4c}<-1+\sqrt{4-4c}<\sqrt{1-4c}\\\left(-1+\sqrt{4-4c}\right)^2<1-4c\\1-2\sqrt{4-4c}+4-4c<1-4c\\-2\sqrt{4-4c}+4<0\\2<\sqrt{4-4c}\\4<4-4c\\c<0.$$ Thus, if $c<0,$ then $\left[-1,-\frac13\right]$ is a subset of the range of $f.$
It remains only to consider $0<c<1,$ in which case $(2)$ and $(3)$ will hold, so that $f$ achieves a local maximum in the interval $(j,k).$ Observe that $$1-c<1\\\sqrt{1-c}<1\\2\sqrt{1-c}<2\\\sqrt{4-4c}<2\\-2+\sqrt{4-4c}<0\\\frac{-2+\sqrt{4-4c}}2<0\\k<0,$$ meaning that $f$ achieves its local maximum at $x=-\sqrt{c},$ and a local minimum at $x=\sqrt{c}$ by the first derivative test. Since $f$ is increasing and continuous on $(-\infty,j),$ and since $\lim_{x\to-\infty}f(x)=1,$ then $f$ is positive on $(-\infty,j).$ Further, since $f\bigl(\sqrt c\bigr)$ is positive by prior work, then since $f$'s minimum value in $(k,\infty)$ occurs at $x=\sqrt c,$ then we have that $f$ is positive on $(k,\infty).$ Thus, we need only ensure that $f\left(-\sqrt c\right)<-\frac13.$ Since $0<c<1,$ then $c<\sqrt c,$ and so the following are equivalent: $$f\left(-\sqrt c\right)<-\frac13\\-3f\left(-\sqrt c\right)>1\\-3\cdot\frac{c-\sqrt c+c}{c-2\sqrt c+c}>1\\-3\cdot\frac{2c-\sqrt c}{2c-2\sqrt c}>1\\-3\left(2c-\sqrt c\right)<2c-2\sqrt c\\-6c+3\sqrt c<2c-2\sqrt c\\5\sqrt c<8c\\\frac58<\sqrt c\\\frac{25}{64}<c.$$
Putting it all together, we find that the range of $f$ does not contain the interval in question if and only if $c>\frac{25}{64}.$
Refer to the Desmos graph.
Find the local maximum: $$\begin{align}f'(x)&= \left(\frac{x^2 + x + c}{x^2 + 2x + c}\right)'=\\ &=\frac{(2x+1)(x^2+2x+c)-(x^2+x+c)(2x+2)}{(x^2 + 2x + c)^2}=0 \Rightarrow \\ x^2&=c \Rightarrow x=\pm \sqrt{c}\end{align}$$ Note the local maximum occurs for $x=-\sqrt{c}$, for which: $$f(-\sqrt{c})=\frac{2\sqrt{c}-1}{2\sqrt{c}-2}<-1 \Rightarrow c\in \left(\frac9{16},1\right).$$ For $c\in [1,+\infty)$: $$f(x)=\frac{x^2+x+c}{x^2+2x+c}=1-\frac{x}{x^2+2x+c}>0.$$ Hence, when $c>\frac9{16}$, the function's range will not contain $[-1,-\frac13]$.
