题目

已知函数f(x)=x^2-2x+3在闭区间[0,m]上的最大值为2m,最小值为2,则m的值是

提问时间：2021-03-26

答案

f(x)=x^2-2x+3
f'(x) =2x-2=0
x=1
f''(x) = 2 >0 (min)
case 1:m ≥1
min f(x) = f(1)
= 1-2+3
= 2
case 1.1:if max f(x) =f(0)= 3
2m=3
m= 3/2 ≥1
accept case 1.1
case 1.2:if max f(x) =f(m)= 3
m^2-2m +3 = 2m
m^2-4m+3 =0
(m-1)(m-3)=0
m=1 or 3
accept case 1.2
case 2:m