※ 引述《jojoba26 (joe)》之銘言：
: 家教學生問我的,我一點頭緒都沒有,請教各位,謝謝!
: 設a,b,c為實數,其中a不為零, 令f(x)=ax^2+bx+c, g(x)=ax+b
: 已知當 |x|<= 1時, |f(x)|<= 5, 試證
: (1)當|x|<= 1時, |g(x)|<= 10
: (2)若已知當 |x|<= 1時,g(x)的最大值為10, 則 f(x)=?
(1)
|c|=|f(0)|<=5
5>=|ax^2+bx+c|>=|ax^2+bx|-|c|=|a|+|b|-|c| for x=1 if ab>0 or x=-1 if ab<0
so |a|+|b|<=|c|+5<=10
|ax+b|<=|a|+|b|<=10 for |x|<=1
(2)
we have |a|+|b|=10
let a>0
it is easy to see that c=-5
so ax^2+bx must be non-negative for |x|<=1
but 0 is a root of ax^2+bx, thus 0 must be a dobule root and ax^2+bx=10x^2
hence f(x)=10x^2-5
the case for a<0 is similar and we can get that f(x)=-10x^2+5
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◆ From: 123.192.216.62
※ 編輯: bineapple 來自: 123.192.216.62 (03/06 12:44)
※ 編輯: bineapple 來自: 123.192.216.62 (03/06 12:45)