设f(x)=lnx+根号x-1证明x>1.f(x)

设f(x)=lnx+根号x-1证明x>1.f(x)
设f(x)=lnx+根号x-1证明x>1.f(x)<3/2(x-1)

设f(x)=lnx+根号x-1证明x>1.f(x)
g(x)=lnx+根号x-1-3/2(x-1)
g(x)=1/x+1/(2√x)-3/2
=(1/x)-1+(1/2)(1/√x-1)
=(1-x)/x+(1/2)(1-√x)/√x
=(1-x)(1/x+(1/2)/[(1+√x)√x])1)
所以g(x)在x>1单减
x>1时,g(x)