一道导数题
f(x)=sinx/(2+cosx)(1)求其单调区间(2)当x大于等于0时,求证f(x)小于等于(1/3)x...
f(x)=sinx/(2+cosx)
(1)求其单调区间
(2)当x大于等于0时,求证f(x)小于等于(1/3)x 展开
(1)求其单调区间
(2)当x大于等于0时,求证f(x)小于等于(1/3)x 展开
1个回答
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(1) f(x)=sinx/(2+cosx) ,f'(x)=(2cosx+1)/(2+cosx) ^2
当2cosx+1>0, 2kπ -2π/3< x < 2kπ +2π/3, f'>0 ,f 增;
当2cosx+1<0, 2kπ +2π/3< x < 2kπ +4π/3, f'<0 ,f 减;
故,单调增区间为[2kπ -2π/3 , 2kπ +2π/3]; 单调减区间为[2kπ +2π/3 , 2kπ +4π/3];
(2)令 g(x)= sinx/(2+cosx) -(1/3)x, x≥0
g'(x) = (2cosx+1)/(2+cosx) ^2 -1 /3
= -(cosx-1)^2 / [3(2+cosx) ^2 ]≤0
当 x≥0,g(x) ≤g(0)=0,
sinx/(2+cosx) -(1/3)x ≤0 得证。
当2cosx+1>0, 2kπ -2π/3< x < 2kπ +2π/3, f'>0 ,f 增;
当2cosx+1<0, 2kπ +2π/3< x < 2kπ +4π/3, f'<0 ,f 减;
故,单调增区间为[2kπ -2π/3 , 2kπ +2π/3]; 单调减区间为[2kπ +2π/3 , 2kπ +4π/3];
(2)令 g(x)= sinx/(2+cosx) -(1/3)x, x≥0
g'(x) = (2cosx+1)/(2+cosx) ^2 -1 /3
= -(cosx-1)^2 / [3(2+cosx) ^2 ]≤0
当 x≥0,g(x) ≤g(0)=0,
sinx/(2+cosx) -(1/3)x ≤0 得证。
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