Question

已知函数f（x）=4x²-4ax+（a²-2a+2）在闭区间【0,2】上有最小值3,求实数a的值

Answer 1

f(x)的对称轴x=a/2,图像开口向上

①当a/2＜0,即a＜0时,最小值为f(0)=a²-2a+2,所以a²-2a+2=3,解得a=1-√2

②当0≤a/2≤2,即0≤a≤4时,最小值为f(a/2)=a²-2a²+a²-2a+2=2a²-4a+2=2(a²-2a+1)=3,解得a=√6/2+1

③当a/2＞2,即a＞4时,最小值为f(2)=16-8a+a²-2a+2=a²-10a+18=3,解得a=√10+5

综上：a=1-√2,a=√6/2+1或a=√10+5