Question

求二次函数f(x)=x*x-2mx+1在区间[-1,1]上的最大值和最小值.

Answer 1

f(x) = x^2-2mx+1 = (x-m)^2 - m^2 + 1

开口向上,对称轴x = m

当m≤-1时,区间在对称轴右侧单调增

最小值f(-1) = 1+2m+1 = 2m+2

最大值f(1) = 1-2m+1 = -2m+2

当m≥1时,区间在对称轴左侧单调减

最大值f(-1) = 1+2m+1 = 2m+2

最小值f(1) = 1-2m+1 = -2m+2

当-1＜m≤0时,

最大值f(1) = 1-2m+1 = -2m+2

最小值 = 极值 = - m^2 + 1

当0≤m＜1时,

最大值f(-1) = 1+2m+1 = 2m+2

最小值 = 极值 = - m^2 + 1