问答题
设f(x)在(a，b)内可导，证明：
【正确答案】正确答案：充分性：设(*)成立，
x
1
，x
2
∈(a，b)且x
1
＜x
2
→ f(x
2
)＜(x
1
)+f'(x
1
)(x
2
—x
1
)，f(x
1
)＜f(x
2
)+f'(x
2
)(x
1
—x
2
)． 两式相加 → [f'(x
1
)一f'(x
2
)](x
2
一x
1
)＞0 → f'(x
1
)＞f'(x
2
)，即f'(x)在(a，b)单调减少． 必要性：设f'(x)在(a，b)单调减少．对于
x
1
，x
2
∈(a，b)且x
1
＜x
2
→ f(x
2
)＜(x
1
)+f'(x
1
)(x
2
—x
1
)，f(x
1
)＜f(x
2
)+f'(x
2
)(x
1
—x
2
)． 两式相加 → [f'(x
1
)一f'(x
2
)](x
2
一x
1
)＞0 → f'(x
1
)＞f'(x
2
)，即f'(x)在(a，b)单调减少． 必要性：设f'(x)在(a，b)单调减少．对于
【答案解析】