Question

请教一下这三道题，麻烦写纸上拍过来，谢了～

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Answer 1

1. 解：

   f( 0 * 0 ) = f(0) * f(0)   ===>    f(0) = 1

   f( x * 1 ) = f(x) * f(1)   ===>    f(1) = 1

   因为

   [ f(1+h) - f(1) ] / h = { f[ h * (1+1/h) - f(1) }  / h =  { f(h) * f(1+1/h) - f(1) }  / h =

   [ f(x + h) - f(x) ] / h = {  f[ x * (1+ h/x) - f(x) ] } / h =   {  [ f(x)* f(1+ h/x) - f(x) ] } / h = 

   =    f(x)* { f(1+ h/x) - 1 ] } / h = [  f(x)/x ] * { f(1+ h/x) - f(1) ] } / (h/x)

       ===>   [  f(x)/x ] * f '(1)

   所以     f '(x) = f '(1) *  [  f(x)/x ]

   解得    f (x) = C * x^[f'(1)]

   f (1) = C * 1^[f'(1)] C = 1    ===>  C = 1

   f(0)  = 1   ===>  f'(1) = 0

   f (x) = 1

2.

f [ g(x) ]  = 1  ,    (  x < 0 )

f [ g(x) ]  = 0  ,    (  x = 0 )

f [ g(x) ]  = - 1  ,    (  x > 0 )

g [ f(x) ] =  e  ,     (  /x/  < 1   )

g [ f(x) ] =  1  ,     (  /x/  = 1   )

g [ f(x) ] =  1/e  ,     (  /x/  > 1   )

 

3.    证明：
a * f(x) +  b * f(1/x) = c/x

a * f(1/x) +  b * f(x) = cx

联解以上两式， 解得

f(x) = [ c/(a^2 - b^2 ) ] * ( a/x - bx )

f(-x) = [ c/(a^2 - b^2 ) ] * ( - a/x + bx ) = - [ c/(a^2 - b^2 ) ] * ( a/x - bx )  = - f(x)

 即    f(-x)= - f(x)