已知函数f(x)=a{2sin(x/2)+sinx}+b,求(1)当a=1时,f(x)的单调递减区间拜托各位了3Q
已知函数f(x)=a{2sin(x/2)+sinx}+b,求(1)当a=1时,求f(x)的单调递减区间(2)当a
彭其美回答:
﹙1﹚当a=1时,f(x)=√2sin(x-π/4)+b+1;则此时的单调递减区间为[2kπ+3π/4,2kπ+7π/4];﹙2﹚∵f(x)=a{2sin(x/2)+sinx}+b=a×√2sin(x-π/4)+b+a∵x∈[0,π],∴x-π/4∈[-π/4,3π/4]∴sin(x-π...
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