I need to find the maxima of the given function: $$f(x)=\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}$$ I have tried to consider both the functions separately and then find their separate maxima, but it didn't work.
I tried to rationalise them into a polynomial fraction, namely: $$f(x)=\frac{-2x^2-6x+12}{\sqrt{x^4-3x^2-6x+13}+\sqrt{x^4-x^2+1}}$$ and use AM-GM method (inequality) for the denominator, but couldn't find a valid solution.
Is there any other way to approach the problem that I am missing?
On
Starting from @user772784's answer we need to solve for $x$ the equation $$\left(3 x^2-x-3\right) \left(4 x^5-12 x^4+2 x^3+3 x^2-5 x-3\right)=0$$ and this is quite discouraging. So, there must be a trick somewhere and I suppose that, at leats, one of the required roots comes from the quadratic, that is to say $x_\pm=\frac{1\pm\sqrt{37}}{6} $
Now, if you plug these value in $f(x)$, the result is $\sqrt{10}$ for the maximum $(x=x_-)$.
Unfortunately, the minimum corresponds to the only real root of the quintic and we cannot get its analytical expression. Using calculus, the minimum is located around $x=2.795$ and its value looks like $-1.547$.
You want to take the derivative of the function and set the derivative to $0$, i.e. you have to solve the equation:
$$f’(x)=0 \tag{1}$$
So, let first find the first derivate of your function $f$:
$$f’(x)=\frac{4x^3-6x-6}{2 \sqrt{x^4-3x^2-6x+13}} -\frac{4x^3-2x} {2\sqrt{x^4-x^2+1}}$$
To solve the equation (1) you need to get rid of the denominators and you’re left with:
$$\sqrt{(x^4-x^2+1)(4x^3-6x-6)^2}-\sqrt{(x^4-3x^2-6x+13)(4x^3-2x)^2}=0$$
Solving that equation will be the hard part.