已知定义在区间[-π,2/3π]上的函数f(x)=Asin(ax+b)(A>0,a>0,-π/2<b<π/2)的图像关于直线x=-π/6对称,当x∈[-π/6,2π/3]时,f(x)的图像如图所示.①求f(x)在[-π,2/3π]上的表达式②求方程f(x
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![已知定义在区间[-π,2/3π]上的函数f(x)=Asin(ax+b)(A>0,a>0,-π/2<b<π/2)的图像关于直线x=-π/6对称,当x∈[-π/6,2π/3]时,f(x)的图像如图所示.①求f(x)在[-π,2/3π]上的表达式②求方程f(x](/uploads/image/z/587080-64-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-%CF%80%2C2%2F3%CF%80%5D%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3DAsin%EF%BC%88ax%2Bb%EF%BC%89%EF%BC%88A%EF%BC%9E0%2Ca%EF%BC%9E0%2C-%CF%80%2F2%EF%BC%9Cb%EF%BC%9C%CF%80%2F2%EF%BC%89%E7%9A%84%E5%9B%BE%E5%83%8F%E5%85%B3%E4%BA%8E%E7%9B%B4%E7%BA%BFx%3D-%CF%80%2F6%E5%AF%B9%E7%A7%B0%2C%E5%BD%93x%E2%88%88%5B-%CF%80%2F6%2C2%CF%80%2F3%5D%E6%97%B6%2Cf%EF%BC%88x%EF%BC%89%E7%9A%84%E5%9B%BE%E5%83%8F%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA.%E2%91%A0%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%9C%A8%5B-%CF%80%2C2%2F3%CF%80%5D%E4%B8%8A%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%E2%91%A1%E6%B1%82%E6%96%B9%E7%A8%8Bf%EF%BC%88x)
已知定义在区间[-π,2/3π]上的函数f(x)=Asin(ax+b)(A>0,a>0,-π/2<b<π/2)的图像关于直线x=-π/6对称,当x∈[-π/6,2π/3]时,f(x)的图像如图所示.①求f(x)在[-π,2/3π]上的表达式②求方程f(x
已知定义在区间[-π,2/3π]上的函数f(x)=Asin(ax+b)(A>0,a>0,-π/2<b<π/2)的图像关于直线x=-π/6对称,当x∈[-π/6,2π/3]时,f(x)的图像如图所示.
①求f(x)在[-π,2/3π]上的表达式
②求方程f(x)=根号2/2的解
已知定义在区间[-π,2/3π]上的函数f(x)=Asin(ax+b)(A>0,a>0,-π/2<b<π/2)的图像关于直线x=-π/6对称,当x∈[-π/6,2π/3]时,f(x)的图像如图所示.①求f(x)在[-π,2/3π]上的表达式②求方程f(x
①观察[-π/6,2π/3]图像可知
A=1 代点得
(π/6)a+b=π/2
(2π/3)a+b=π
解得a=1,b=π/3
∴在[-π/6,2π/3]上,f(x)=sin(x+π/3)
又∵其关于x=-π/6对称
∴在[-π,-π/6]上,f(x)=sin[2×(-π/6)-x+π/3]=-sinx
∴f(x)={sin(x+π/3)x∈[-π/6,2π/3]
-sinx x∈[-π,-π/6] 此乃分段函数.
②此可理解为画y=√2/2与图像交点问题.
令-sinx=√2/2
解得x=-π/4或-3π/4 在区间里
再令sin(x+π/3)=√2/2
解得x=-π/12或5π/12
综上,方程f(x)=√2/2,x∈[-π,2/3π]
的解集为{-π/4,-3π/4,-π/12,5π/12}
【补充】那个关于直线对称,求解析式问题有结论:
函数y=f(x)关于直线x=a对称的解析式为f(2a-x)
例如f(x)=x²-2x+1关于直线x=1对称的函数为什么?
用2a-x替换每一个x,即(2-x)²-2(2-x)+1
图在哪里?