<s>[INST] Minima and maxima of a trig function let $f(x) = \sqrt {4\sin^4x - \sin^2x\cos^2x + 4\cos^4x}$ 
what is the product of the minimum and maximum value of the function?
I tried turning them all to $\sin x$
$-\sin^2x\cos^2x = \sin^4x-\sin^2x ; 4\cos^4x = (1-\sin^2x)^2 = 4sin^4x - 4sin^2x +4 $
which gives us :
$9\sin^4x - 2\sin^2x +4$
then i dont know how to proceed
EDIT: 
The equation above is wrong, i got if you try solving for the ones i have given : 
$9\sin^4x - 5\sin^2x +4$
then I tried making $sin^2x = a$ (Thanks @Greninja for the info)
where in the the min is when $a = 5/18$ and the value is 47/36, 
then the max is when $a=1$ and the value is 8 
Min x Max = $12\sqrt{94}$ which looks far fetched
I dont know where I went wrong with my solution Please help.
And yes, I have been told that the answer is $\sqrt{7}$
 [/INST] First I wrote 
$$4 \sin^4 x+4 \cos^2 x-\sin^2 x \cos^4 x=9 \cos ^4 x-9 \cos ^2 x+4$$
Then remembering that $\cos 4x =8 \cos ^4 x-8 \cos ^2 x+1$
I got 
$$f(x)=\sqrt{4 \sin ^4 x-\sin ^2 x \cos ^2 x +4 \cos ^4 x}= \frac{\sqrt{9 \cos 4 x+23}}{2 \sqrt{2}}$$
$$f'(x)=-\frac{9 \sin 4 x}{\sqrt{2} \sqrt{9 \cos 4 x+23}}$$
and 
$$f''(x)=-\frac{9 (92 \cos 4 x+9 (\cos 8 x+3))}{\sqrt{2} (9 \cos 4 x+23)^{3/2}}$$
Setting $f'(x)=0$ I got $\sin 4x=0\to 4x=k\pi\to x=+k\pi/4,\;k\in\mathbb{Z}$
Plugging in the second derivative
$f''(\pi/4)=\frac{18}{\sqrt{7}}>0$ therefore $x=\pi/4$ is a point of minimum where $f(\pi/4)=\frac{\sqrt{7}}{2}$
$f''(0)=-9/2$ so $x=0$ is a maximum and $f(0)=2$
So the product is $2\frac{\sqrt{7}}{2}=\sqrt{7}$
Hope this can be useful
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