(2012新課標9)已知$\omega>0,$函式$f(x)=sin(\omega x+\dfrac{\pi}{4})$在$(\dfrac{\pi}{2},\pi)$上單調遞減,則$\omega$的取值範圍是______

分析:

常規方法:$\dfrac{\pi}{2}+2k\pi\le\omega x+\dfrac{\pi}{4}\le\dfrac{3\pi}{2}+2k\pi,k\in Z$
得$x\in[\dfrac{\pi+8k\pi}{4\omega},\dfrac{5\pi+8k\pi}{4\omega}]$取$k=0$得$\dfrac{\pi}{4\omega}\le\dfrac{\pi}{2},\dfrac{5\pi}{4\omega}\ge\pi$得$\omega\in[\dfrac{1}{2},\dfrac{5}{4}]$
巧法:利用影象伸縮變換，如圖

先對函式$f(x)=\sin(x+\dfrac{\pi}{4})$作圖,$f(x)=\sin(\omega x+\dfrac{\pi}{4})$是由上圖縱座標不變，橫座標伸縮為原來的$\dfrac{1}{\omega}$所得.

考慮$\dfrac{\pi}{4\omega}=\dfrac{\pi}{2},\dfrac{5\pi}{4\omega}=\pi$易得$\omega\in[\dfrac{1}{2},\dfrac{5}{4}]$