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!
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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.
Hint: suppose $\,c, x \gt 0\,$, then $\,u = x + \dfrac{c}{x} \ge 2 \sqrt{c}\,$ by AM-GM with equality iff $\,x = \sqrt{c}\,$, and:
$$ f(x)=\frac{x^2-3x+c}{x^2+3x+c} \color{red}{\cdot \frac{\;\;\cfrac{1}{x}\;\;}{\cfrac{1}{x}}}=\dfrac{x+ \cfrac{c}{x}-3}{x+ \cfrac{c}{x}+3} = \dfrac{u-3}{u+3}=1 - \cfrac{6}{u+3} \ge 1 - \frac{6}{2 \sqrt{c}+3} $$
The RHS gives a lower bound, which is actually a minimum since it is attained for $\,x = \sqrt{c}\,$.
For it to equal $\,\dfrac{1}{7}\,$, it follows that $\displaystyle\,1 - \frac{6}{2 \sqrt{c}+3}=\frac{1}{7} \iff 2 \sqrt{c} +3 = 7 \iff c = 4\,$.
What remains to be filled in is the details to cover the rest of cases, which are fairly straightforward to work out, for example $\,f(-x) = \dfrac{1}{f(x)}\,$.