已知函数y=tan(2x+α)的图像过点(π/12,0)则函数y=tan(2x+α)的单调区间为

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已知函数y=tan(2x+α)的图像过点(π/12,0)则函数y=tan(2x+α)的单调区间为

已知函数y=tan(2x+α)的图像过点(π/12,0)则函数y=tan(2x+α)的单调区间为
已知函数y=tan(2x+α)的图像过点(π/12,0)则函数y=tan(2x+α)的单调区间为

已知函数y=tan(2x+α)的图像过点(π/12,0)则函数y=tan(2x+α)的单调区间为
y=tan(2x+α)的图像过点(π/12,0)
所以 0=tan(π/6+α)
α=-π/6
y=tan(2x-π/6)
值域增区间
kπ-π/2

由题意得,tan(2x π/12+α)=0 得α=-π/6, 则 πk-π/2≤2x-π/6≤πk+π/2
kπ/2-π/6,kπ/2+π/3

tan(π/6+α)=0,π/6+α=kπ,α=kπ-π/6
y=tan(2x+kπ-π/6)=tan(2x-π/6)
kπ-π/2<2x-π/6y=tan(2x+α)的单调区间为[kπ/2-π/6,kπ/2+π/3]