$y = \sqrt{1 + cos(4x^2)} - 2$.
$y' = \dfrac{-4x sin(4x^2)}{\sqrt{1 + cos(4x^2)}}$
y'=0 <-> x=0 hoac $x=\dfrac{\sqrt{k \pi}}{2}$
Thay vao ham so ban dau ta co
$y(0) = \sqrt{2}-2$, $y(\dfrac{\sqrt{k \pi}}{2}) = -2$ neu k le va $y(\dfrac{\sqrt{k \pi}}{2}) = \sqrt{2}-2$ neu k chan.
Vay GTNN la -2 voi $x=\dfrac{\sqrt{k \pi}}{2}$ voi k le va GTLN la $\sqrt{2}-2$ tai x=0 hoac $x=\dfrac{\sqrt{k \pi}}{2}$ voi k chan.
