Calculate the Extrema of $ G(ω) = \frac{|15(1-ω^2)|}{\sqrt{25(3-ω^2)^2 +ω^2(11-ω^2)^2}} $ I want to calculate this without the use of software. I also want to find the points of Inflection but this might be too hard by hand, so let's forget that for now. So i know what is the procedure in order to find maxima and minima. I calculated the first derivative of G(ω) and i found: 
$$ \frac{d}{dω}G(\omega) = \frac{30\omega\ sgn(\omega^2-1)\ \sqrt{25(3-\omega^2)^2 +\omega^2(11-\omega^2)^2} +\frac{50\omega|15(1-\omega^2)|(3-\omega^2)}{\sqrt{25(3-\omega^2)^2 +\omega^2(11-\omega^2)^2}} + 2\omega (11-\omega^2)(11-3\omega^2)}{25(3-\omega^2)^2 +\omega^2(11-\omega^2)^2} $$
$\forall ω>0$, where $sgn(x)$ is the signum function
Now i have to find the roots of this. Those points ($\omega$) will be the extrema. So by equating to $0$ i get:$$ 15\ sgn(ω^2-1)\ \big[ 25(3-ω^2)^2+ω^2(11-ω^2)^2\big] +25|15(1-ω^2)|(3-ω^2) + (11-ω^2)(11-3ω^2)\sqrt{25(3-ω^2)^2+ω^2(11-ω^2)^2} = 0 $$ How can i solve this for $ω$? I put this equation in Mathcad software and it outputs:"No solution was found"... Is there any way to solve this? How to get rid of $sgn(x)$? If anyone could provide any help i would be very grateful. Thanks in advance.
UPDATE
Thanks to Tony Piccolo's and mickep's comments below, if i make the substitution $Ω=ω^2$ and taking the derivative of $G(Ω)^2$, i seem to get the correct results without much trouble. The following is a graph i did on Mathcad. It draws the graph, but it can't find the roots of the derivative (the extrema).


Here are some thoughts, too long for a comment.
Firstly, the expression is even in $\omega$ (which I will write $w$), so it is sufficient to consider $w\geq 0$.
Secondly, it is clear that the expression is non-negative, and that it is zero precisely when $w=1$. Thus, that is the smallest value.
Thirdly, instead of studying this complicated expression, it is easier to consider the square of it. The good thing is that it will have the same max/min (squared of course). Thus, I recommend you to study
$$
f(w)=G(w)^2=\frac{\bigl(15(1-w^2)\bigr)^2}{25(3-w^2)^2+w^2(11-w^2)^2}.
$$
Differentiating, you end up at solving a polynomial equation of rather high degree (six?). I don't know if it is possible to solve it with square roots...
