finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-10 x-7 x^2+x^4}}+\frac{1+3x-2x^3}{\sqrt{5-2 x-3 x^2+x^4}}$$.
But failed to solve $f'(x)=0$ for finding $f(x)_{max}$.
I would be glad for your help.
Thanks.

Hint: 

Consider the points $A=(4,5)$ and $B=(1,2)$ and a point $P$ on the line $x=y$ parallel to the line AB. We have for the distances $PA$, $PB$ $AB$ the triangle inequality 
$$PA-PB \le AB$$ with equality if an only if $A$,$B$, $P$ lie on a line in this order. This will only happen for points $P$ chosen on $x=y$ at $-\infty$. The supremum value is $AB = 3\sqrt{2}$, the maximum is not achieved.
