pure (differentiation) urgent

用戶90065

2010-01-27 6:23 am

(a) Given that y=f(x) is a function satisfies the equation dy/dx = ky, where k is a constant. Let u(x) = ye^-kx. By differentiating u(x) with respect to x, show that y=ce^kx for some constant c.
(b) Let f(x) be a non-constant function defined on R such that
(1) f(x+y) = f(x)f(y)
(2) f(x)=1+xg(x)
(3) lim g(x) =1
x->0
Prove that (i) f'(x) exists for all x
(ii) f(x)=e^x

回答 (1)

Prof. Physics

2010-01-27 8:14 am

✔ 最佳答案

a. du/dx = d(ye^-kx)/dx = yd(e^-kx)/dx + e^-kx dy/dx

= -kye^-kx + kye^-kx

= 0

So, u = c, where c is a constant

ye^-kx = c

y = ce^kx

b.i. Consider lim @x→0 [f(x + @x) - f(x)]/(x + @x - x)

= lim @x→0 [f(@x)f(x) - f(x)]/@x

= lim @x→0 f(x)(f(@x) - 1)/@x

= lim @x→0 f(x)[(@x)g(@x)]/@x

= lim @x→0 f(x)g(@x)

= f(x) lim @x→0 g(@x)

= f(x)

So, f'(x) exists for all x.

ii. f'(x) = f(x) for all x

From part a, k = 1

So, f(x) = ce^x, where c is a constant

However, from condition (2):

ce^x = 1 + xg(x)

lim x→0 ce^x = lim x→0 (1 + xg(x))

c = 1 + lim x→0 xlim x→0 g(x)

c = 1 + 0 = 1

So, f(x) = e^x

參考: Physics king

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