<s>[INST] 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!
 [/INST] 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 $\,f(-x) = \dfrac{1}{f(x)}\,$.
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