calculus
posted by farouk on .
Show that the equation;x+3sinx=2 has a root between x=0.4 and x=0.6 using newton raphson's formular only

f(x) = x + 3sinx  2
f'(x) = 1 + 3cosx
1: f(0.6000000) = 0.2939274
2: f(0.5154411) = 0.0058029
3: f(0.5170484) = 0.0000019
4: f(0.5170490) = 0.0000000 
Farouk, It would be a good idea to check back the previous posts you have made before reposting. It may save you a long wait.
See answer to your previous post:
http://www.jiskha.com/display.cgi?id=1321390404