Question

已知函數(shù)f(x)=e^x-ln(x+1)-1(x≥0).

(Ⅰ)求f(x)的最小值；

(Ⅱ)若x＞y≥0，求證：e^x-y-1＞ln(x+1)-ln(y+1).

Answer 1

解：(Ⅰ)∵f(x)=e^x-ln(x+1)-1∴f′(x)=e^x 

∵x≥0,∴e^x≥1,≤1∴e^x≥0,即f′(x)≥0 

∴f(x)=e^x-ln(x+1)-1在[0,+∞]是增函數(shù) 

∴x=0時(shí),f(x)有最小值,最小值為f(0)=0 

(Ⅱ)∵x＞y≥0,∴x-y＞0根據(jù)(Ⅰ)f(x)=ex-ln(x+1)-1在[0,+∞]是增函數(shù)

∴f(x-y)＞f(0),即e^x-y-ln(x-y+1)-1＞0 

∴e^x-y-1＞1n(x-y+1) 

要證e^x-y-1＞ln(x+1)-ln(y+1),只需證ln(x-y+1)＜ln(x+1)-ln(y+1)

即證ln(x-y+1)＞ln,只需能證x-y+1＞

∵x＞y≥0,∴x-y+1=＞0

∴e^x-y-1＞ln(x-y+1)＞ln(x+1)-1n(y+1) 

即e^x-y-1＞ln(x+1)-ln(y+1)。