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第五次習題 Exercise 4.4-28
- 1. 學號:U10011024 姓名:王冠倫
Exercise 4.4-28
2 − 2x − x2 , −2 ≤ x ≤ 0
f (x) = |x − 2| , 0<x<3
1 2
3 (x − 2) , 3≤x≤4
因為
−2 − 2x
, −2<x<0
−1 , 0<x<2
f (x) =
1 , 2<x<3
2
(x − 2)2
3 , 3<x<4
令f (x) = 0, f (x)不存在, f (x)不連續,得駐點x1 =
−1, x2 = 0, x3 = 2, x4 = 3,又因為當−2<x< −
1, 2<x<3, 3<x<4時,f (x)>0;當−1<x<0, 0<x<2時,
f (x)<0。列表討論如下:
x −2 (−2, −1) −1 (−1, 0) 0
f (x) + 0 − 不存在
f (x) 2 3 2
x (0, 2) 2 (2, 3) 3 (3, 4) 4
f (x) − 不存在 + 1 +
1 8
f (x) 0 3 3
其中對任何x ∈ U (−1, 1),有f (x)<f (−1),知當x =
−1時,函數有極大值f (−1) = 3;對任何x ∈ U (0, 1),
1
- 2. 沒有f (x)<f (0)或f (x)>f (0),知當x = 0時,函數沒
有極值;對任何x ∈ U (2, 1),有f (x)>f (2),又對任
何x ∈ U (3, 1),有f (x)>f (3),知當x = 2, 3時,函數
1
有極小值f (2) = 0, f (3) = 3 ;對任何x ∈ U (0, 1), x>−
2,有f (x)>f (−2),知當x = −2時,函數有endpt
minf (−2) = 2;對任何x ∈ U (4, 1), x<4,有f (x)<f (4),
知當x = 4時,函數有endpt maxf (4) = 8 。而f (−1)>f (x),
3
f (2)<f (x),知當x = −1時,函數有最大值與極大
值f (−1) = 3,當x = 2時,函數有最小值與極小值
f (2) = 0。是故,
f (−2) = 2 endpt min , f (−1) = 3 最小值,極小值
f (0) 沒有極值 , f (2) = 0 最大值,極大值
1
f (3) = 3 極小值 , f (4) = 8 endpt max
3
2