一個問題 數學 想很久 - 拼圖
By Ida
at 2009-06-12T18:29
at 2009-06-12T18:29
Table of Contents
※ 引述《jerrylibra (GO)》之銘言:
: 請數學好的人幫 幫我
: 這一題我算了好久
: 題目是 有一個2次函數 y=x^2+4ax+b
: 然後有一個區間 -1<x<1 然後對應 -1<y<2
: 求 a,b是多少?
y'' = 2 > 0
y' = 2x+4a
所以y的極小值-1出現在y'=0(即x=-2a) or x=-1 or x=1 處
將y(x)分4個狀況討論
case1: y(-1)=-1, y(1)=2
case2: y(1)=-1, y(-1)=2
case3: y(-2a)=-1, y(-1)=2
case4: y(-2a)=-1, y(1)=2
***********************分析分隔線***********************
case1: a=3/8, b=-1/2 ok
case2: a=-3/8, b=-1/2 ok
case3: y(-2a)=-4a^2+b=-1
y(-1)=1-4a+b=2
可得a=1/2+√(3)/2, b=3+2√3 →-2a < -1 不合
或 a=1/2-√(3)/2, b=3-2√3 ok
case4: y(-2a)=-4a^2+b=-1
y(1)=1+4a+b=2
可得a=-1/2+√(3)/2, b=3-2√3 ok
或 a=-1/2-√(3)/2, b=3+2√3 →-2a > 1 不合
共得到以上四種可能的解
--
: 請數學好的人幫 幫我
: 這一題我算了好久
: 題目是 有一個2次函數 y=x^2+4ax+b
: 然後有一個區間 -1<x<1 然後對應 -1<y<2
: 求 a,b是多少?
y'' = 2 > 0
y' = 2x+4a
所以y的極小值-1出現在y'=0(即x=-2a) or x=-1 or x=1 處
將y(x)分4個狀況討論
case1: y(-1)=-1, y(1)=2
case2: y(1)=-1, y(-1)=2
case3: y(-2a)=-1, y(-1)=2
case4: y(-2a)=-1, y(1)=2
***********************分析分隔線***********************
case1: a=3/8, b=-1/2 ok
case2: a=-3/8, b=-1/2 ok
case3: y(-2a)=-4a^2+b=-1
y(-1)=1-4a+b=2
可得a=1/2+√(3)/2, b=3+2√3 →-2a < -1 不合
或 a=1/2-√(3)/2, b=3-2√3 ok
case4: y(-2a)=-4a^2+b=-1
y(1)=1+4a+b=2
可得a=-1/2+√(3)/2, b=3-2√3 ok
或 a=-1/2-√(3)/2, b=3+2√3 →-2a > 1 不合
共得到以上四種可能的解
--
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