標題:
Maths f(x)
發問:
Let f(x) be a continuous function for all real values of x such that (1) f(0) is not equal to 0 (2) f '(0)=2 (3) f(a+b)=f(a)f(b) for all real no. a & b Required: (a) Show that f(0)=1 (b) Show that f '(x)=2f(x) for all real values of x 更新: But for (b), if differentiation is not required / cannot be used, can you just solve by using functions and derivative method? 更新 2: Thanks lot!
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最佳解答:
(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。
其他解答:
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.
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