Question

Ai giải giúp em âu này di ạ

Answer 1

\(f'\left(x\right)=2sin\left(x+\dfrac{\pi}{4}\right)cos\left(x+\dfrac{\pi}{4}\right)-sinx=sin\left(2x+\dfrac{\pi}{2}\right)-sinx=cos2x-sinx\)

\(\Rightarrow f'\left(\dfrac{\pi}{4}\right)=cos\left(\dfrac{\pi}{2}\right)-sin\left(\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

b.

\(f'\left(x\right)=0\Leftrightarrow cos2x-sinx=0\)

\(\Leftrightarrow cos2x=sinx=cos\left(\dfrac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}-x+k2\pi\\2x=x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)