f(x)=3x^2+x-2,-3≤x≤0,求f(x)的範圍。

2011-12-29 5:50 pm
f(x)=3x^2+x-2,-3≤x≤0,求f(x)的範圍。
更新1:

答案為-3≤f(x)≤22

更新2:

答案範圍最小是-3

回答 (2)

2011-12-29 6:42 pm
✔ 最佳答案
f(x)=3x^2+x-2,-3≤x≤0,求f(x)的範圍。Sol: 利用(a+b)^2=(a^2+2ab+b^2) 的公式f(x)=3x^2+x-2 =3[x^2+(1/3)x+(1/6)^2]-3*(1/6)^2-2 =3(x+1/6)^2-3/36-2=3(x+1/6)^2-25/12所以當x=-1/6 時有最小值 = -25/12f(0) = 0+0-2=-2f(-3)=3*(-3)^2+(-3)-2 =27-3-2 =22所以最大值為22所以 -25/12≤f(x)≤22

2011-12-29 10:44:26 補充:
解題技巧: 將3x^2+x-2變化成 3(x+a)^2+b 就可以求極值

2011-12-29 15:33:07 補充:
範圍最小是-25/24 -3 是錯誤的答案
參考: 電腦週邊急救團
2011-12-29 7:07 pm
3x^2+x-2
=3(x^2+x/3)-2
=3(x^2+2*x*1/6+1/36)-2-1/12
=3(x+1/6)^2-25/12
-3<=x<=0
-3+1/6<=x+1/6<=1/6
-17/6<=x+1/6<=1/6
0<=(x+1/6)^2<=289/36
0<=3(x+1/6)^2<=289/12
-25/12<=3(x+1/6)^2-25/12<=22
-25/12<=f(x)<=22


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