作者victor7935 (victor)
看板trans_math
标题[微分] 习题...
时间Sun Nov 9 19:07:32 2008
1.
If f'(X)=0 at x=1,x=2,x=3,then f cannot possibly increase on[0,4].
这是对的!!...简单证明
2.
Show that cosx < 1 - (1/2)x^2 + (1/24)x^4 for all x >0.
3.
Show that the equation P(x)=0 has exactly two real roots,
both positive.
P(x)= x^4 - 8x^3 + 22x^2 - 24x + 4
4.Prove that a polynomial of degree n has at most n-1 local extreme values.
5.Let f be differentiable on an open interval I.Prove that,if f'(x) = 0
for all x in I,then f is constant on I.
感觉.....
这些问题感觉对大家来说都很简单!?
有时候都不知道该不该问了= = ....
自己好像还想得不够多....
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※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 140.123.221.147
1F：→ Qmmm:第1题叙述是对的~ 140.112.128.93 11/09 19:33
2F：→ victor7935:= = '我还以为是故意写这样的说!!140.123.221.147 11/09 19:45
3F：→ victor7935:那我修一下好了..140.123.221.147 11/09 19:47
※ 编辑: victor7935 来自: 140.123.221.147 (11/09 19:47)
4F：→ Qmmm:第2题 令f(t)=cost-1+(1/2)t^2+(1/24)t^4 140.112.128.93 11/09 19:50
5F：→ Qmmm:由1阶导数判别 x=0时为绝对极小值 140.112.128.93 11/09 19:51
6F：→ Qmmm:sorry 令f(x)=cosx-1+(1/2)x^2+(1/24)x^4 140.112.128.93 11/09 19:51
7F：→ Qmmm:所以f(x) > 0 for x > 0 ...就证出来了 140.112.128.93 11/09 19:52
8F：→ Qmmm:第三题一样是一阶导数可以作出来 140.112.128.93 11/09 19:54
※ 编辑: victor7935 来自: 140.123.221.147 (11/09 20:11)
9F：→ zptdaniel:第一题..随便抓一个分段定义函数举反例 123.194.99.216 11/09 20:40
10F：→ zptdaniel:domf=[0,4] not [1,3] 123.194.99.216 11/09 20:41
11F：→ zptdaniel:increasing not strictly increasing 123.194.99.216 11/09 20:41