一道高数证明题,急急急,一定会有好评?

2024-10-30 21:22:31
推荐回答(4个)
回答1:

记f(x)=(1+x)ln(1+x)-arctanx, 则当x>0时,f'=ln(1+x)+1-1/(1+x^2)>0,f(x)在(0,正无穷)增,又当x趋于0时,f(x)趋近于0,所以f(x)>0,即(1+x)ln(1+x)>arctanx.

回答2:

本题考查介质定理和拉格朗日中值定理!
∵1/3,2/3∈(0,1)
f(x)在[0,1]上连续,
∴根据介值定理,∃x1,x2∈(0,1),使得:
f(x1)=1/3
f(x2)=2/3
又∵
f(x)在区间(0,x1),(x1,x2),(x2,1)可导,在[0,x1],[x1,x2],[x2,1]连续,
根据拉格朗日中值定理:
∃ξ1∈(0,x1)
∃ξ2∈(x1,x2)
∃ξ3∈(x2,1)
使得:
f(x1)-f(0) =f'(ξ1)·(x1-0)
f(x2)-f(x1)=f'(ξ2)·(x2-x1)
f(1)-f(x2)=f'(ξ3)·(1-x2)
因此:
1/f'(ξ1) = (x1-0)/f(x1)-f(0) =x1/(1/3)=3x1
1/f'(ξ2) = (x2-x1)/f(x2)-f(x1) =(x2-x1)/(1/3)=3x2-3x1
1/f'(ξ3) = (1-x2)/f(1)-f(x2) =(1-x2)/(1/3)=3-3x2
上述各式相加:
1/f'(ξ1) + 1/f'(ξ2) + 1/f'(ξ3) = 3x1+3x2-3x1+3-3x2=3

回答3:

设f(x)=(1+x)ln(1+x)-arctanx
f'(x)=ln(1+x)+1-1/(1+x²)=ln(1+x)+x²/(1+x²)
当x>0时,ln(1+x)>0,x²/(1+x²)>0
因此f(x)在(0,+∞)上单增
因此对任意x>0,总有f(x)>f(0)=0
即(1+x)ln(1+x)>arctanx

回答4:

。。。。。。。。

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