<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] Solving $$y=\frac{x^2-3x+c}{x^2+3x+c}$$ and $$y=7$$
we put $$\frac{x^2-3x+c}{x^2+3x+c}=7$$
to get $$x^2-4x+c=0$$ which should have exactly one root.
So putting $Discriminant=0$
we get $c=4$
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