View source for Page 536 problem 6 MA181Fall2008bell - Rhea

<math>g(x+y) = \frac{g(x)+g(y)}{1-g(x)g(y)}</math>

<math>\lim_{h \to 0} g(h) = 0</math>

<math>\lim_{h \to 0} \frac{g(h)}{h}= 1</math>

a. Show that <math> g(0) = 0</math>.

b. Show that <math>g'(x) = 1 + [g(x)]^2</math>

c. Find <math>g(x)</math> by solving the differential equation in part (b).

----

Anyone know where to start?  I'm defeated at every turn; I can't break the function into even/odd portion that have any use and none of the laws of exponentials/logarithms seem to be very useful.  The only fact I can pull out is that <math>g'(0)=1</math> which can be determined through L'Hopitals.

--[[User:Jmason|Jmason]] 15:28, 5 October 2008 (UTC)


* You can show g(0) = 0 by solving for g(x) (Yes, you can do it.  No, it's not that hard), and then plugging 0 in for x.  As for the other parts, I haven't got that far yet.  I'll see what I get.  And wow, I've been working on this problem a half hour already, I think.[[User:Jhunsber|Jhunsber]]