Question

已知f(x)=4x²-4ax+a²-2a+2在区间[0,2]上的最小值为3，求a取值。

过程和答案，谢谢

Answer 1

由题设，f(x)=4x²-4ax+a²-2a+2，知f(x)图像开口向上，且在区间[0,2]上为单增函数，f(0)=a²-2a+2，在x轴上方，且在y轴上。
因此，f(x)min=f(0)=a²-2a+2=3
推出a=1±根号2

Answer 2

f(x)=4x²-4ax+a²-2a+2
f'(x) = 8x-4a =0
x= a/2
f''(x) = 8 >0 (min)

case 1: a/2<0
min f(x) = f(0)
= a^2-2a+2 = 3
a^2-2a-1=0
a= 1-√2 or 1+√2
solution for case 1 : a= 1-√2

case 2: 0 ≤a/2≤2
min f(x) =f(a/2)
= a^2 - 2a^2+a^2-2a+2
= -2a+2 = 3
a=-1/2
no solution for case 2:

case 3: a/2>2
min f(x) = f(2)
= 16-8a+a²-2a+2
=a^2-10a+18=3
a^2-10a+15=0
a =5+√10 or 5-√10
solution for case 3: a =5+√10

a取值: case 1 or case 2 or case 3
: a=1-√2 or 5+√10

Answer 3

轴动区间定