求二次函數(shù)f(x)=x*x-2mx+1在區(qū)間[-1,1]上的最大值和最小值.

數(shù)學(xué)人氣：526 ℃時(shí)間：2019-11-04 23:31:44

優(yōu)質(zhì)解答

f(x) = x^2-2mx+1 = (x-m)^2 - m^2 + 1
開口向上,對稱軸x = m
當(dāng)m≤-1時(shí),區(qū)間在對稱軸右側(cè)單調(diào)增
最小值f(-1) = 1+2m+1 = 2m+2
最大值f(1) = 1-2m+1 = -2m+2
當(dāng)m≥1時(shí),區(qū)間在對稱軸左側(cè)單調(diào)減
最大值f(-1) = 1+2m+1 = 2m+2
最小值f(1) = 1-2m+1 = -2m+2
當(dāng)-1＜m≤0時(shí),
最大值f(1) = 1-2m+1 = -2m+2
最小值 = 極值 = - m^2 + 1
當(dāng)0≤m＜1時(shí),
最大值f(-1) = 1+2m+1 = 2m+2
最小值 = 極值 = - m^2 + 1