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)
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