已知m为实数,f(x)=x^2-2mx+m-1的最小值是f(m),试求函数f(m)在0≤m≤2上的最大值最小值,
问题描述：
已知m为实数,f(x)=x^2-2mx+m-1的最小值是f(m),试求函数f(m)在0≤m≤2上的最大值最小值,
答
f(x) = x^2-2mx+ m-1
f'(x) = 2x - 2m = 0
x = m
f''(x) = 2 > 0 ( min)
minf(x)= f(m) = m^2-2m^2+ m-1
= -m^2+m-1
= -(m-1/2)^2 - 3/4
maxf(m) = f(1/2) = -3/4
minf(m) = f(2) = -(2-1/2)^2 - 3/4
= -9/4 - 3/4
= -3
答
f(x)=(x-m)²-m²+m-1
所以最小值f(m)=-m²+m-1
=-m²+m-1/4+1/4-1
=-(m-1/2)²-3/4
对称轴m=1/2,开口向下
0≤m≤2
所以m=2,最小值=-3
m=1/2,最大值=-3/4