Find $m$ and $n$ so that the given function has the range $[-3, 5]$ Find $m$ and $n$ real numbers so that $f(x) = \frac{3x^2 + mx + n}{x^2 + 1}$ takes all and only the values from the interval $[-3, 5]$.
I started by solving the following double inequality:
$$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$$
From the left inequality I got $6x^2 + mx + n + 3 \geq 0$. The coefficient of $x^2$ is positive so the discriminant $\delta$ has to be less than or equal to $0$.
$\delta = m^2 - 24n - 72 \leq 0$
From the right inequality I got $2x^2 -mx + 5 - n \geq 0$. By using the same technique as the one used to solve the left inequality I got:
$\delta = m^2 - 40 + 8n \leq 0$
By combining these two resulted inequalities I got $m \leq 48$, that is $m \in [-4\sqrt3, 4\sqrt3]$.
Now I need to find the values of $m$ and $n$ so that $f$ is surjective, because it must take all the values in the given interval.
I don't know yet how to proceed, so I would appreciate any help from you guys!
Thank you!
 A: $\frac{3x^2 + mx + n}{x^2 + 1}$ reaching a maximum value of $5$ implies that $5(x^2+1)-3x^2+mx+n = 2x^2-mx-(n-5)$ has a double root.  Similarly, $\frac{3x^2 + mx + n}{x^2 + 1}$ reaching a minimum value of $3$ implies that $3x^2+mx+n-(-3)(x^2+1) = 6x^2+mx+(n+3)$ has a double root.  Quadratic expressions have a double root when their determinant $b^2-4ac = 0$, so we obtain the simultaneous equations
$$
m^2+8(n-5) = 0
$$
and
$$
m^2-24(n+3) = 0
$$
which yields $n = -1$ and $m = \pm 4\sqrt{3}$.  From Desmos:

A: Here's another way. Consider $\large y=\dfrac{3x^2+mx+n}{x^2+1}$. Rearranging, we have:
$$x^2(y-3)+-mx+(y-n)=0$$
Now, since $x$ has to be real (since domain of $y$ is $\mathbb{R}$), so we see that:
$$m^2 - 4(y-3)(y-n) \ge 0 \\
  \implies 4y^2-4(n+3)y+12n-m^2 \le 0 $$
Since $y$ is surjective, so the entire range of $y$ must be exactly $[-3,5]$, so our equation should be equivalent to $(y+3)(y-5)\le 0$, or:
$$4y^2-8y-60 \le 0$$
So we just compare the coefficients to get the two equations:
$$-4(n+3)=-8 \\
 12n-m^2= -60$$
Which gives:
$$ m=\pm 4\sqrt 3 \text{ and } n= -1$$
A: Background:  A parabola graph of $f(x)=ax^2 + bx + c$ will have it's maximum/minimum value at its vertex, $x = -\frac b{2a}$.
Review: because $f(x) = a(x + \frac b{2a})^2 - a(\frac b{2a})^2 +c$.  As $(x + \frac b{2a})^2 \ge 0$ with equality holding only if $x = - \frac b{2a}$, $f(x)$ achieves it's max min value of $- \frac {b^2}{4a}+c$ and $x =- \frac b{2a}$.
$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$ 
(and achieves points where equality holds; for the rest of this post, if I write $F(x) \ge a$ I'm going to take it to mean both $F(x) \ge a$ and there exists some (at least one) $y$ so that $F(y) = a$.)
so $-3x^2 - 3 \le 3x^2 + mx + n$ 
so $0 \le 6x^2 + mx + n+3$
So $6x^2 + mx + (n+3)$ acheives its minimum at $x = -\frac {m}{12}$ and the minimum value is $-\frac{m^2}{24} + n + 3 = 0$
Also
$-3 \leq \frac{3x^2 + mx + n}{x^2 + 1} \leq 5$
so $3x^2 + mx + n \le 5x^2 + 5$ 
so $0 \le 2x^2 - mx - n+5$
So $2x^2 - mx - n+5$ acheives its minimum at $x = \frac {m}{4}$ and the minimum value is $-\frac{m^2}{8} - n + 5 = 0$
So it becomes a matter of solving for $m,n$ where $-\frac{m^2}{8} - n + 5 = 0$ and $-\frac{m^2}{24} + n + 3 = 0$
So $n= 5 - \frac{m^2}{8} = \frac{m^2}{24} - 3$
$\frac{m^2}{24}+\frac{m^2}{8}=8$
$4m^2 = 8*24$
$m = \pm \sqrt{48} = \pm4\sqrt{3}$
$n = 5 - 6 = 2 - 3 = - 1$
Thank's to FreezingFire for pointing out the type that led to a numerically incorrect result.
