How to find the max value of $\frac{x^4 − 3x^3 − 11x^2 + 3x + 10}{3 − x^2}$ with $-1 \leq x \leq 1$ I'm trying to find the max value of this function in this interval. I now that in order to do that I have to find the value where this functions derivative is equal to zero, and with the second derivative I will know if it is a maximum or a minimum value.
$f(x) = \frac{x^4 − 3x^3 − 11x^2 + 3x + 10}{3 − x^2}$
$f'(x) = \frac{-2x^5+3x^4+12x^3-24x^2-46x+9}{\left(3-x^2\right)^2}$
I know that if the derivative of $f'_1 =-2x^5+3x^4+12x^3-24x^2-46x+9 = 0$ is already enough to me. But this is where I'm stuck, I don't really know how to solve this.
 A: As you wrote
$$f(x)=\frac{x^4-3 x^3-11 x^2+3 x+10}{3-x^2}$$
$$f'(x)= \frac{-2x^5+3x^4+12x^3-24x^2-46x+9}{\left(3-x^2\right)^2}$$ If we want to avoid purely numerical calculations, let us think about approximations.
First of all, notice that we have
$$f(-1)=0 \qquad \text{and} \qquad f'(-1)=+6$$
$$f(1)=0 \qquad \text{and} \qquad f'(1)=-12$$
With such a conformation, there is a maximum which is not at any bound.
On another side
$$f(0)=\frac{10}{3}\qquad \text{and} \qquad f'(0)=1$$ Considering how much did change the first derivative, its zero must not be very far from $x=0$.
So, expanding the derivative around $x=0$ as a Taylor series, we have
$$f'(x)=1-\frac{46 }{9}x-2 x^2+O\left(x^3\right)$$ Its closest solution to zero is
$$x_*=\frac{\sqrt{691}-23}{18} \sim 0.182604$$ Checking
$$f'(x_*)=\frac{7763038-296231 \sqrt{691}}{1698939}\sim -0.0140973$$ So, consider that $x_*$ is almost the maximum point
$$f(x_*)=\frac{4714 \sqrt{691}-51209}{21222} \sim \color{blue}{3.42604} $$
This does not seem too bad since a complete numerical work would give a maximum value of $\color{red}{3.42605}$ (!!!) at $x=0.180284$.
Edit
For the fun of it, adding one more term to the Taylor series
$$f'(x)=1-\frac{46 }{9}x-2 x^2-\frac{56 }{27}x^3+O\left(x^4\right)$$ The cubic shows only one real root. Using the hyperbolic method, it is given by
$$x_*=\frac{1}{28} \left(2 \sqrt{563} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{13257}{563 \sqrt{563}}\right)\right)-9\right)\sim 0.180514$$ which is much better.
Now, computing
$f(x_*)=\color{red}{3.426052}70$ while the exact value is $\color{red}{3.42605286}$.
