祥佑 吳 asked in 科學數學 · 8 years ago

能幫我解這題微分問題嗎

題組題:x及y為實數,則滿足f(xy)=f(x)+f(y)

1.求f(1)

2.求證f(x)=-f(1/x)

3.若f '(1)=a,求 f '(a)

想請問第三題之解及過程

1 Answer

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  • 8 years ago
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    @前言

    考慮f在0的值,對於任一實數x皆有

    f(0) = f(0*x) = f(0)+ f(x) ---> f(x) =0 ,因為對於任意實數x,f(x)皆為0故f(x)為零函數

    我們考慮non-trivail case,假設該函數在0點未定義,則我們考慮非零點:

    1.

    f(1) = f(1*1) = f(1)+f(1) = 2*f(1) ---> f(1) = 0

    2.

    令x不等於0 ,由1.得知

    0 = f(1) = f(x*(1/x)) = f(x)+f(1/x) ---> f(x) = -f(1/x)

    3.

    設f為C1實數函數, a不等於0

    f(x) = f(x/a)+f(a)

    f '(x) = f '(x/a)/a (連鎖律且f(a)為一常數)

    令x=a

    f '(a) = f '(1)/a = a/a = 1

    2012-07-31 14:45:43 補充:

    更正:第四行non-trivial

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