If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?
I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appreciated!

Solution
First, we should constrain the value of $c$ such that $x^2+3x+c \neq 0$ for all $x \in \mathbb{R}.$ Otherwise, there necessarily exists at least one infinite discontinuity for $y=f(x)$, and if so, there exists no minimum or maximum value for $y$. For this purpose, let $\Delta=9-4c<0$, i.e.    $c>\dfrac{9}{4},$ which is enough.
Under this constraint, $y=f(x)$ is continuous over $(-\infty,+\infty)$. It's clear that the maximum and minimum value given can only be reached at the local extremum point.
Now,notice that $$y'=f'(x)=\frac{6(x^2-c)}{(x^2+3x+c)^2}.$$Let $y'=0$. Then $$x=\pm\sqrt{c}.$$
Hence,$$f(\sqrt{c})=\frac{2\sqrt{c}-3}{2\sqrt{c}+3}=1-\frac{6}{2\sqrt{c}+3}<1,~~~~~f(-\sqrt{c})=\frac{2\sqrt{c}+3}{2\sqrt{c}-3}=1+\frac{6}{2\sqrt{c}-3}.$$
Hence, $f(\sqrt{c})=\dfrac{1}{7},$ then $$c=4.$$
Finally,we may verify that $c=4$ could satisfy all the conditions. We are done.
