Question

y = \(2-3\sqrt{5+2cos\left(2x+\dfrac{\pi}{4}\right)}\)

Answer 1

Tìm min, max

 

Answer 2

\(-1\le cos\left(2x+\dfrac{\pi}{4}\right)\le1\Rightarrow-2\le2cos\left(2x+\dfrac{\pi}{4}\right)\le2\)

\(\Rightarrow\sqrt{3}\le\sqrt{5+2cos\left(2x+\dfrac{\pi}{4}\right)}\le\sqrt{7}\)

\(\Rightarrow2-3\sqrt{7}\le y\le2-3\sqrt{3}\)

\(y_{max}=2-3\sqrt{3}\) ; \(y_{min}=2-3\sqrt{7}\)

Answer 3

Answer 4

\(y=2-3\sqrt{5+2cos\left(2x+\dfrac{\pi}{4}\right)}\ge2-3\sqrt{5+2.1}=2-3\sqrt{7}\)

\(y=2-3\sqrt{5+2cos\left(2x+\dfrac{\pi}{4}\right)}\le2-3\sqrt{5+2.\left(-1\right)}=2-3\sqrt{3}\)

\(\Rightarrow min=2-3\sqrt{7}\Leftrightarrow cos\left(2x+\dfrac{\pi}{4}\right)=1\Leftrightarrow x=-\dfrac{\pi}{8}+k\pi\)

\(max=2-3\sqrt{3}\Leftrightarrow cos\left(2x+\dfrac{\pi}{4}\right)=-1\Leftrightarrow x=\dfrac{3\pi}{8}+k\pi\)