设函数f(x)=arctanx,g(x)=x,x＞0
f(0)=0,g(0)=0
f'(x)=1/(1+x²)＞0,g'(x)=1＞0
f'(x)-g'(x)=1/(1+x²)-1=-x²/(1+x²)≤0即f'(x)≤g'(x)
f(x)与g(x)在左端点处的函数值相同,在[0,+∞)上f(x)与g(x)单调递增且f'(x)≤g'(x),所以有arctanx≤x,仅当x=0时取等号
设函数f(x)=x,g(x)=ln(1+x),x＞0
f(0)=0,g(0)=0
f'(x)=1＞0,g'(x)=1/(1+x)＞0
f'(x)-g'(x)=1-1/(1+x)=x/(1+x)＞0即f'(x)＞g'(x)
f(x)与g(x)在左端点处的函数值值相同,在(0,+∞)上f(x)与g(x)单调递增且f'(x)＞g'(x),所以有x＞ln(1+x)