设f(x)=lnx+x^1⼀2-1,证明:当x>1时f(x)<3⼀2(x-1)

2025-02-22 22:45:38
推荐回答(1个)
回答1:

lnx<=x-1 记t=根号x
f<3/2(x-1)
x-1+t-1<3/2(x-1)
2(t-1)<(x-1)
2t4x<(x+1)^2
0<(x-1)^2