Question

若函数f(x)在【0,1】上满足f”(x)>0,则f‘(0)

设在区间[0,1]上f''(x)>0,则f'(0)f'(1)和f(1)-f(0)的大小顺序是
设在区间[0,1]上f"(x)>0,则f'(0)、f'(1)和f(1)-f(0)的大小顺序是:
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)

Answer 1

因为f(x)在[0,1]上一阶可导,由Lagrange中值定理,f(1)-f(0) = f'(ξ)(1-0)=f'(ξ).其中ξ∈[0,1],又由于f''(x)>0 => f'(x)在[0,1]上为单调递增函数,于是有f'(1) > f(1)-f(0)=f'(ξ) > f'(0)