Maths f(x)

(a) From (3), put a＝b＝0, getsf(0＋0)＝f(0) f(0)==> [f(0)]²－f(0)＝0==> f(0) [f(0)－1]＝0==> f(0)＝0 (reject, from (1)) or f(0)＝1∴ f(0)＝1
(b) Differentiate both sides of (b) w.r.t. x, getsf'(a＋b) d(a＋b)/dx＝f(a) f'(b) d(a)/dx＋f(b) f'(a) d(b)/dxput a＝0, b＝x, getsf'(0＋x) d(0＋x)/dx＝f(0) f'(x) d(0)/dx＋f(x) f'(0) d(x)/dx==> f'(x) * 1＝1＊f(x)＊0＋f(x)＊2＊1 ⋯⋯⋯⋯ (dx/dx＝1，d(0)/dx＝0，from (2), f'(0)＝2)==> f'(x)＝2f(x)

2014-10-01 20:48:19 補充：
What do you mean? Differentiation is the derivative method.
What's mean differentiation cannot be used but must solve by derivative method?

2014-10-01 23:05:45 補充：
原來你説的是 derive from first principle,
f'(x)
＝lim(h→0) [f(x＋h)－f(x)]/h
＝lim(h→0) [f(x) f(h)－f(x)]/h
＝f(x) lim(h→0) [f(h)－1]/h
Put x＝0, gets
f'(0)＝f(0) lim(h→0) [f(h)－1]/h
==> 2＝1＊lim(h→0) [f(h)－1]/h ⋯⋯ from (2) and part (a)
∴ lim(h→0) [f(h)－1]/h＝2
So, f'(x)
＝f(x) lim(h→0) [f(h)－1]/h
＝f(x)＊2
＝2f(x)

2014-10-01 23:06:35 補充：
感謝 貓Sir 的提供。

Bryan，其實你呢題題係係好 common 的以前的 Pure Math 的練習題。

的確在證明 f(x) 是 differentiable everywhere 之前直接用 d/dx 呢個 operator 唔係太好。

你應該用 first principle 這樣做：
先由 (2) 證明 lim(h→0)[ f(h) - 1 ]/h = 2
Put a = 0 and b = h in (3), consider
f'(0) = 2
lim(h→0)[f(0 + h) - f(0)]/h = 2
lim(h→0)[f(0)f(h) - f(0)]/h = 2

2014-10-01 21:13:09 補充：
f(0) lim(h→0)[f(h) - 1]/h = 2
lim(h→0)[f(h) - 1]/h = 2

然後，考慮 for any x,
Consider
　lim(h→0)[f(x + h) - f(x)]/h
= lim(h→0)[f(x)f(h) - f(x)]/h
= f(x)lim(h→0)[f(h) - 1]/h
= f(x)(2)
= 2f(x) is well defined.

Therefore, f'(x) = 2f(x) for all x.