作者Honor1984 (希望愿望成真)
看板trans_math
标题Re: [积分] 90政大资管
时间Thu Jul 10 12:52:22 2008
※ 引述《t90766 (Feng)》之铭言：
: x
: Let f(x)=∫ e^[(t-1)/t^2] dt, x≧1. Find the interval(s)
: 1
: where f is concave upward, and the interval(s) where f is concave downward.
: 微了一次还是e...要怎麽解呀= =???
: 先感谢各路高手!!!
考虑x≧1的情况
f'(x) = e^[(x-1)/x^2]
f''(x) = e^[(x-1)/x^2] * [x^2-(x-1)*2x]/x^4
= e^[(x-1)/x^2] * [-x^2+2x]/x^4
= e^[(x-1)/x^2] * [-x^2+2x]/x^4
= e^[(x-1)/x^2] * [-(x)(x-2)]/x^4
=> f''(x)<0 for x>2 -------------concave downward
f''(x)>0 for 2>x≧1 ----------------concave upward
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◆ From: 122.124.96.6
1F：推 t90766:谢谢你! 这题目告诉我坚持下去答案就会出来220.132.175.253 07/10 13:28
2F：→ Honor1984:对,要有耐性,慢慢写 122.124.96.6 07/10 14:51