f(x)=（2sin(x+π/6)+x^4+x）/(x^4+cosx))+1则有f(x)=2sin(x+π/6)+x^4+x/(x^4+cosx)则有在闭区间-π/2到闭区间π/2的最大值最小值为M与N,求证M+N=4?

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f(x)=（2sin(x+π/6)+x^4+x）/(x^4+cosx))+1则有f(x)=2sin(x+π/6)+x^4+x/(x^4+cosx)则有在闭区间-π/2到闭区间π/2的最大值最小值为M与N,求证M+N=4?
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f(x)=（2sin(x+π/6)+x^4+x）/(x^4+cosx))+1则有f(x)=2sin(x+π/6)+x^4+x/(x^4+cosx)则有在闭区间-π/2到闭区间π/2的最大值最小值为M与N,求证M+N=4?
f(x)=（2sin(x+π/6)+x^4+x）/(x^4+cosx))+1则有
f(x)=2sin(x+π/6)+x^4+x/(x^4+cosx)则有在闭区间-π/2到闭区间π/2的最大值最小值为M与N,求证M+N=4?

f(x)=（2sin(x+π/6)+x^4+x）/(x^4+cosx))+1则有f(x)=2sin(x+π/6)+x^4+x/(x^4+cosx)则有在闭区间-π/2到闭区间π/2的最大值最小值为M与N,求证M+N=4?
先化简肯定没问题,然后再观察式子的特点.
∵2sin(x+π/6)=√3sinx+cosx
∴f(x)=(2sin(x+π/6)+x^4+x)/(x^4+cosx)+1
=(√3sinx+cosx+x^4+x)/(x^4+cosx))+1
=(√3sinx+x)/(x^4+cosx)+2
设g(x)=(√3sinx+x)/(x^4+cosx)
在区间[-π/2,π/2]上,g(-x)=-g(x)
即g(x)在[-π/2,π/2]上奇函数,也就是说函数图像关于原点对称.
设g(x)在[-π/2,π/2]的最大值和最小值分别是T与t
由于g(x)关于原点对称,所以T+t=0
而f(x)=g(x)+2在[-π/2,π/2]的最大值M=T+2,最小值m=t+2
∴M+N=(T+2)+(t+2)=(T+t)+(2+2)=4

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