Max and min application

$\begin{aligned} y& = \frac {x^2-3x+c}{x^2+3x+c} \\ & = \frac {x^2+3x+c - 6x}{x^2+3x+c} \\ & = 1 - \frac {6x}{x^2+3x+c} \\ & = 1 - \frac {6}{x+3+\dfrac cx} \end{aligned}$

We note that $y$ is maximum when $x+3+\dfrac cx$ is least negative , that is $\left|x+3+\dfrac cx\right|$ is minimum but with a negative sign, and minimum when $x+3+\dfrac cx$ is is least positive , that is $\left|x+3+\dfrac cx\right|$ is minimum with a positive sign. In both case we need only to find $\left|x+3+\dfrac cx\right|$ and we can consider both $x$ and $\dfrac cx$ as positive and apply AM-GM inequality as: $x + \dfrac cx \ge 2\sqrt c$ and equality occurs when $x = \sqrt c$ .

Therefore, when maximum and minimum occur, $y = 1 - \dfrac {6}{x+3+\dfrac cx} = 1 - \dfrac 6{\pm 2\sqrt c + 3}$ and that, when:

$\implies \begin{cases} y = 7 & \implies -2\sqrt c + 3 = - 1 & \implies \sqrt c = 2 & \implies c = \boxed{4} \\ y = \frac 17 & \implies 2\sqrt c + 3 = 7 & \implies \sqrt c = 2 & \implies c = \boxed{4} \end{cases}$

$y = \dfrac {x^2-3x+c}{x^2+3x+c} \\ =1 - \dfrac {6x}{x^2+3x+c} \\ \therefore\ \ \dfrac{dy}{dx}= - 6*(x^2+3x+c)+6x*(2x+3)=0.\\ \implies x^2+3x+c =2x^2+3x.\\ \implies x^2 =c\\ \color{#3D99F6}{x=\pm \sqrt c.} \\ Substituting\ - \sqrt c \ for\ x, \\ y_{max}=1- \dfrac{6\sqrt c}{c+3\sqrt c +c}=7\\ \implies\ \ \sqrt c =2c - 3*\sqrt c\\ \therefore \ \ \sqrt c=2\\ \Huge\ \ \ \ \color{#D61F06}{c=4.}\\ Putting\ \sqrt c = x = 2\\ y_{min}=1- \dfrac{ 6* 2}{4+ 3* 2 +4}=\dfrac 1 7.$

A calculus approach:

$y=\dfrac{x^2-3x+c}{x^2+3x+c}\\ \dfrac{dy}{dx}=\dfrac{(x^2+3x+c)(2x-3)-(x^2-3x+c)(2x+3)}{(x^2+3x+c)^2}\\ =\dfrac{x^3+6x^2+2cx-3x^2-9x-3c-(x^3-6x^2+2cx+3x^2-9x+3c)}{(x^2+3x+c)^2}\\ =\dfrac{6x^2-6c}{(x^2+3x+c)^2}$

At maximum, minimum points, $\dfrac{dy}{dx}=0$

$\dfrac{6x^2-6c}{(x^2+3x+c)^2}=0\\ 6x^2-6c=0\\ x^2=c\\ x=\pm\sqrt{c}$

Substitute $x^2=c$ and the maximum value of $y$ :

$7=\dfrac{c-3x+c}{c+3x+c}\\ 7(2c+3x)=2c-3x\\ 14c+21x=2c-3x\\ 12c+24x=0\\ c+2x=0$

Now, we have two values of $x$ here, so we try both:

$x=\sqrt{c}\\ \implies c+2\sqrt{c}=0\\ \sqrt{c}(\sqrt{c}+2)=0\\ c=0,\;\sqrt{c}=-2$

The second factor has no real roots, therefore we ignore it.

Note that if $c=0,\;x=0,\;x^2+3x+c=0$ . This will result in the denominator becoming $0$ , so we reject this as well.

Try the other value of $x$ :

$x=-\sqrt{c}\\ \implies c-2\sqrt{c}=0\\ \sqrt{c}(\sqrt{c}-2)=0\\ c=0,\;\sqrt{c}=2\implies c=4$

We rejected $c=0$ , so we know that $c=4$ . As a precaution, check back the values:

$c=4,\;x=2,\;y=\dfrac{4-6+4}{4+6+4}=\dfrac{2}{14}=\dfrac{1}{7}\\ c=4,\;x=-2,\;y=\dfrac{4+6+4}{4-6+4}=\dfrac{14}{2}=7$

Therefore, our answer is $c=\boxed{4}$

Note: You can try out the second half of the solution with $y=\dfrac{1}{7}$ instead. You will still get the two same equations, and the answer will remain the same.

For a more rigorous approach, you can also show that the values are maximum and minimum with the second derivative test