2个回答
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令x=0,y=0,则f(0+0)=f(0)*f(0)
f(0)=f(0)^2
f(0)*[f(0)-1]=0
因为f(x)是非零函数,所以f(0)≠0,即f(0)=1
f'(x)=lim(t->0) [f(x+t)-f(x)]/t
=lim(t->0) [f(x)*f(t)-f(x)]/t
=f(x)*lim(t->0) [f(t)-1]/t
=f(x)*lim(t->0) [f(0+t)-f(0)]/t
=f(x)*f'(0)
=f(x)*1
=f(x)
f(0)=f(0)^2
f(0)*[f(0)-1]=0
因为f(x)是非零函数,所以f(0)≠0,即f(0)=1
f'(x)=lim(t->0) [f(x+t)-f(x)]/t
=lim(t->0) [f(x)*f(t)-f(x)]/t
=f(x)*lim(t->0) [f(t)-1]/t
=f(x)*lim(t->0) [f(0+t)-f(0)]/t
=f(x)*f'(0)
=f(x)*1
=f(x)
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