# 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!

• Just checking that you are expected to do this without any principles from calculus? Sep 30 '16 at 18:18
• Well, I found this exercise in a precalculus book, but I would appreciate any solution :) Sep 30 '16 at 18:21

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: $$m=\pm 4\sqrt 3 \text{ and } n= -1$$

$\frac{3x^2 + mx + n}{x^2 + 1}$ 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:

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.

• The problem is, the equation is not a parabola. It is just the quotient of two quadratic polynomials. So this method won't work! Oct 1 '16 at 5:05
• Doesn't matter! I referred to three DIFFERENT functions 1) f (x)= (3x^+mx+n)/(x^2+1); 2) g (x)= 6x^2 + mx + n +3; and 3) h (x) =2x^2- mx - n +5. I noted $-3 \le f (x) \iff g (x) \ge 0$ and that $f (x) \le 5 \iff h (x) \ge 0$. And g (x) and h (x) most certainly ARE parabolas. So yes, this method will work. Reread what I wrote. I never used f(x) directly to find the the values of m and n. Oct 1 '16 at 6:58
• I am guilty of not reading your answer carefully, and I'm sorry for that! Thank you for pointing that out. Just your second equation needs correction: $-(m^2/8)-n+\color{blue}{5}=0$. Just one doubt, why must the minimum value of $g(x)$ be exactly zero? Why not $\ge 0$? Rest is clear! Oct 1 '16 at 12:58
• That "5" probably explains where my arithmetic error is. why must the minimum of $g(x) \ge 0$ be exactly $0$. Well, I guess technically it doesn't. After all $x^2 \ge -3,547$ but the minimum is $0$ and not $-3,547$. But in context I was using and taking $g(x) \ge 0$ to mean $g(x) \ge 0$ and that there exisis an x somewhere such that $g(x) = 0$. This is, I believe, acceptable as the question states not only that $-3 \le f(x) \le 5$ but that it achieves all points. If this is true then $0\le g(x)$ AND $g(x)= 0$ for some $x$. If there were a symbol for $f(x)\ge a;\exists x f(x)=a$. I'd use it. Oct 1 '16 at 16:07