作者LuisSantos (^______^)
看板trans_math
標題Re: [考古] 交大８９
時間Sat Jul 7 17:22:25 2007
※ 引述《johnnyzsefb (AJ)》之銘言：
: 3.(10 points)
: x^2
: Let f:R→R be the function given by f(x)=∫(2-e^{t^2}) dt
: -x^2
: Find the maximum of f if it exists.
: 我把f對x微分，得到在x=0 or x=(ln2/lnt)^(1/4) 有極值，但是我就不會接下來的
: 積分了(∫e^{t^2}dt)......
: 以上
: 謝謝
x^2
f(x) = ∫ (2 - e^(t^2)) dt
-(x^2)
f'(x) = (2x)(2 - e^(x^4)) - (-2x)(2 - e^(x^4))
= (2x)(2 - e^(x^4)) + (2x)(2 - e^(x^4))
= (4x)(2 - e^(x^4))
令 f'(x) = 0
則 x = 0 或 2 - e^(x^4) = 0
e^(x^4) = 2
x^4 = ln2 => x^4 - ln2 = 0
=> (x^2 + (ln2)^(1/2))(x^2 - (ln2)^(1/2)) = 0
=> x^2 - (ln2)^(1/2) = 0 (x^2 + (ln2)^(1/2) > 0)
=> x^2 = (ln2)^(1/2)
=> x = (ln2)^(1/4) , -((ln2)^(1/4))
↗ | ↘ ↘ |↗ ↗ |↘
-----------------------------------------------
-((ln2)^(1/4)) 0 (ln2)^(1/4)
所以當 x = (ln2)^(1/4) , -((ln2)^(1/4)) 時 ,
(ln2)^(1/2)
f(x) 有最大值 ∫ (2 - e^(t^2)) dt
-((ln2)^(1/2))
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