推 t90766:謝謝你! 這題目告訴我堅持下去答案就會出來220.132.175.253 07/10 13:28
→ Honor1984:對,要有耐性,慢慢寫 122.124.96.6 07/10 14:51
※ 引述《t90766 (Feng)》之銘言：
: 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.
: 微了一次還是e...要怎麼解呀= =???
: 先感謝各路高手!!!
考慮x≧1的情況
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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