Question

【数学函数单调性】用定义证明f(x)=In(1+e^x)+x在R上单调递增

Answer 1

令x10

f(x2)-f(x1)

=ln(1+e^x2)+x2-ln(1+e^x1)-x1

=ln[(1+e^x2)/(1+e^x1)]+(x2-x1)

因为e^x是增函数,所以e^x2>e^x1

1+e^x2>1+e^x1>0

(1+e^x2)/(1+e^x1)>1

ln x也是增函数

ln[(1+e^x2)/(1+e^x1)]>ln 1=0

所以f(x2)-f(x1)>x2-x1>0

所以f(x)在R上单调增