Exponentiate Negatively, Get Value (Posted on 2007-12-24)

     y
   ∫(g(p))-2 dp = g(y), 
    0
and:
     4
   ∫(g(p))-2 dp = (12)1/3.
    0

	Submitted by K Sengupta
Rating: 3.0000 (1 votes)
Solution:	(Hide)
g(72) = 6 EXPLANATION: Differentiating both sides of the first relationship with respect to y, we obtain: g’(y) = (g(y))-2, so that: (g(y))2*g’(y) = 1 Integrating both sides w.r.t y, we obtain: (g(y))3/3 = y+c Or, (g(y))3 = 3(y+c)……..(*) From the first relationship, we also note that: g(0) = 0, so that, (*) gives: (g(0))3 = 3c, so that: c=0 Accordingly, (g(y))3 = 3y, so that: (g(72))3 = 3*72=216 or, g(72)= 6 *** I am grateful to Chesca Ciprian, who has rightly pointed out that the second condition of the problem is superfluous.

	Subject	Author	Date
	re: Observation..........	K Sengupta	2008-03-06 04:35:54
	re: Observation..........	Kurious	2007-12-26 08:05:19
	Observation..........	Chesca Ciprian	2007-12-25 12:49:56
	re(2): complete solution	Daniel	2007-12-25 10:52:58
	re: complete solution	Kurious	2007-12-25 10:12:40
	complete solution	Daniel	2007-12-25 02:14:37