Question

连续与可导的问题若f(x+1)=af(x)总成立,且f'(0)=b,a,b为非零常数,则f(x)在x=1a .不可导 b

Answer 1

f(t)=af(t-1),

因为t=1时,f(t-1)=f(0)可导,且t-1可导

则 f '(t)=af '(t-1)(t-1)'=af '(0)*1=ab 可导.