设函数f(x)=sin(2x+φ)(-π<φ<0),y=f(x)图像的一条对称轴是直线x=π/8.(1)求φ; (2)画出函数f(x)(2)画出函数f(x)在区间[0,π]上的图像.
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![设函数f(x)=sin(2x+φ)(-π<φ<0),y=f(x)图像的一条对称轴是直线x=π/8.(1)求φ; (2)画出函数f(x)(2)画出函数f(x)在区间[0,π]上的图像.](/uploads/image/z/2485088-8-8.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%86%29%28-%CF%80%EF%BC%9C%CF%86%EF%BC%9C0%EF%BC%89%2Cy%3Df%28x%29%E5%9B%BE%E5%83%8F%E7%9A%84%E4%B8%80%E6%9D%A1%E5%AF%B9%E7%A7%B0%E8%BD%B4%E6%98%AF%E7%9B%B4%E7%BA%BFx%3D%CF%80%2F8.%281%29%E6%B1%82%CF%86%3B+%EF%BC%882%EF%BC%89%E7%94%BB%E5%87%BA%E5%87%BD%E6%95%B0f%28x%29%EF%BC%882%EF%BC%89%E7%94%BB%E5%87%BA%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C%CF%80%5D%E4%B8%8A%E7%9A%84%E5%9B%BE%E5%83%8F.)
设函数f(x)=sin(2x+φ)(-π<φ<0),y=f(x)图像的一条对称轴是直线x=π/8.(1)求φ; (2)画出函数f(x)(2)画出函数f(x)在区间[0,π]上的图像.
设函数f(x)=sin(2x+φ)(-π<φ<0),y=f(x)图像的一条对称轴是直线x=π/8.(1)求φ; (2)画出函数f(x)
(2)画出函数f(x)在区间[0,π]上的图像.
设函数f(x)=sin(2x+φ)(-π<φ<0),y=f(x)图像的一条对称轴是直线x=π/8.(1)求φ; (2)画出函数f(x)(2)画出函数f(x)在区间[0,π]上的图像.
(1)y=f(x)图像的一条对称轴是直线x=π/8,则y=f(x)在x=π/8取得最(大或小)值,即sin(π/4+φ)=±1
而-π<φ<0
所以-3π/4<π/4+φ<π/4
所以sin(π/4+φ)=1不成立,只有sin(π/4+φ)=-1
此时π/4+φ=-π/2即φ=-3π/4
(2)画出函数f(x)如图所示
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