1. If f(x)=(x-3)^2+(x-4)^2, find the minimum value of f(x).
ans:1/2
2. if the maximum value of f(x)=2kx^2-12x-k, which k<0, is 11, find the value of k.
ans:-9 or 2
f.4 a maths 一問.
2007-11-26 1:53 am
回答 (2)
2007-11-26 2:16 am
✔ 最佳答案
f(x)=(x-3)^2+(x-4)^2=x^2-6x+9+x^2-8x+16
=2x^2-14x+25
=2(x^2-7x)+25
=2[x^2-7x+(7/2)^2-(7/2)^2]+25
=2(x-7/2)^2+25-49/2
when x=7/2, f(x)is the smallest.
so the minimum value of f(x)=1/2.
參考: me...
2007-11-26 2:46 am
2. 2kx^2 -12x -k =11 ( x at this point , f(x) is maximum, so there is only one real root x)
2kx^2-12x -(k+11) =0
b^2 -4ac =0 (there is only one real root x as this point, f(x) is maximum)
12^2 - 4(2k) (-k-11) =0
144 +8k^2+ 88k =0
8k^2 +88k +144 =0
k^2 +11k + 18 =0
(k+9)(k+2)=0
k=-9 or -2
2kx^2-12x -(k+11) =0
b^2 -4ac =0 (there is only one real root x as this point, f(x) is maximum)
12^2 - 4(2k) (-k-11) =0
144 +8k^2+ 88k =0
8k^2 +88k +144 =0
k^2 +11k + 18 =0
(k+9)(k+2)=0
k=-9 or -2
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