Question

已知f(x)=-4x^2+4ax-4a-a^2在区间【0,1】内的最大值为-5，求a 的值

Answer 1

f(x)=-4x^2+4ax-4a-a^2
x=-4a/-8=a/2
1)a/2<0 a<0
f(x)max=f(1)=-4+4a-4a-a^2=-5
-4-a^2=-5
a=-1
2)0≤a/2≤1 0≤a≤2
f(x)max=f(a/2)=)=-a²+4a(a/2)-4a-a^2=-5
a=5/4
3)a/2>1 a>2
f(x)max=f(0)=-4a-a^2=-5
a=1 a=-5
∴a=-1 a=5/4

Answer 2

f(x)=-4x^2+4ax-4a-a^2
f'(x) = -8x+4a
f'(x) =0
=> x = a/2
f''(x) = -8 ( max )
if a/2 在区间[0,1]
f(a/2) = -5
-a^2+2a^2-4a-a^2 =5
-4a= 5
a= -5/4 ( rejected)

ie a/2在区间 (-无限,0)U(1,+无限)
=> a在区间 (-无限,0)U(2,+无限)
if a在区间 (-无限,0)
f'(x) = -8x+4a <0 ( decreasing )
max f(x) at x= 0
f(0) = -4a-a^2 =5
a/62+4a-5 =0
(a+5)(a-1) =0
a = -5 or 1 ( rejected)

if a在区间 (2,+无限)
f'(x) = -8x+4a >0 ( increasing )
max f(x) at x= 1
f(1) = -4+4a-4a-a^2 = -5
a^2-1 =0
a= 1 or -1 (rejected )

ie a=1 or a= -5

Answer 3

推荐回答明显错掉了不要误导好吧，只有发图的答案是对的

Answer 4

绝对正确，