函数在0到1的闭区间内二阶导数大于0选择:a.f'(1)>f'(0)>f(1)—f(0)

问题描述:

函数在0到1的闭区间内二阶导数大于0选择:a.f'(1)>f'(0)>f(1)—f(0)
b.f'(1)>f(1)—f(0)>f'(0)
c.f(1)—f(0)>f'(1)>f'(0)
d.f'(1)>f(0)—f(1)>f'(0)

函数f(x)在0到1的闭区间内二阶导数大于0
表明函数f(x)在0到1的闭区间内一阶导数f'(x)是增函数,显然f'(1)>f'(0)
且函数f(x)在0到1的闭区间内是(向上)凹函数
f(1)-f(0)=[f(1)-f(0)]/(1-0)=k
是过端点A(0,f(0))和B(1,f(1))的弦AB的斜率k,
k等于与弦AB平行且与这段抛物线弧相切的切线斜率,因此,它介于f'(1)与f'(0)之间.Lagrange中值定理可以说明这一点.
选B