已知x>1,求证:x-(x²/2)-lnx-1\2
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已知x>1,求证:x-(x²/2)-lnx-1\2
已知x>1,求证:x-(x²/2)-lnx-1\2
已知x>1,求证:x-(x²/2)-lnx-1\2
证明:令f(x)=x-(x²/2)-lnx
f(x)'=1-x-1/x
当x>1时,f(x)'<0
所以f(x)=x-(x²/2)-lnx在(1,+无穷)单调递减
即f(x)<f(1)
(1)=1-1/2-0
=1/2
所以f(x)<1/2
即可知:x-(x²/2)-lnx-1\2