已知函数fx=sinxcosx+1–2sinx平方,x属于r 求函数y=fx的对称轴方程和单调递

区间

f(x) = 2sinxcosx+1-2sin²x
= sin2x + cos2x
= √2(sin2xcosπ/4+cos2xsinπ/4)
= √2sin(2x+π/4)
2x+π/4=2kπ±π/2时取极值,此时,x+π/8=kπ±π/4
∴对称轴为x=kπ-3π/8,x=kπ+π/8,其中k∈Z

单调递增区间:(kπ-3π/8,kπ+π/8);单调递减区间:(kπ+π/8,kπ+5π/8);其中k∈Z
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