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ROOTS OF EQUATIONS	 GRAPHIC METHOD
GRAPHIC METHOD This method is used primarily for find an interval where function has  any root. Tounderstandbetterwegoingto do anExample. fx=arctan(x)+(x−1)  
solution To findtherood of f(x), we do arctanx+x−1=0,  wherewehavearctanx=1. Thus, theproblemistofindthepoint of intersection of thegraphs of thefunctionsgx=arctanx , y, hx=1−x. So, wegraphthis.  
Graphictofindtherood g(x)= arctan x h(x)= 1-x Wherewe can seecrearlythatanintervalwheretherootisonliinterval (0,1)
Someconsiderations If fa∗fb<0,  is probably to find an odd number of roots for this equation.   If fa∗fb>0,  is probably to find an even number of roots or that there aren’t.  
OPEN METHODS	 FIXED POINT
FIXED POINT This methodisappliedtosolveequations of theformx=gx Iftheequationisfx=0, thenyou can eithercleared x oradded x in bothsides of theequationtoput in theproperly.   CLOSE METHODS
Graphicallywehave.. f1x=x , f2=g(x)   y=f(x) Raíz Iteratively x1+1=g(xi)   - CLOSE METHODS
Someconditionstoconvergence If f2(x)≤f1x  y  f2x>0:  we’llhave a monotonicallyconvergentsolutionbecauseeachsolutionisobtainedclosertotheroot. If f2(x)≤f1x  y  f2x<0:  it has a convergentoscillatorysolutionbecauseeachsolutionisobtained in a mannerclosertotherootoscillatory. Iff2(x)≥f1x  y  f2x>0,  so it has a divergentsolution.   CLOSE METHODS
example We needfindtherootsforthisequiation: fx=e−x−lnx So gx=e−x−lnx+x then you have to do a table like:   Fixedpoint Tolerance = 0.01
CLOSE METHODS	 NEWTON-RAPHSON
Newton-Raphson This method, which is an iterative method is one of the most used and effective.  Newton-Raphson method does not work on a range bases his formula in an iterative process.Suppose we have the approximation xi to the root xr of   f(x),   f(x) tangente this line intersects the axis x,  at a point xi+1that will be our next approximation to the root xr.   Xi Xr Xi+1
To calculatedthepointxi+1, first we have to find equation of the tangent line. We know that id has pending m=f′(xi) So, theequationis: y−fxi=f′xix−xi Afterwe do Y=0 −fxi=f′xix−xi And solvefor x: x=xi−f(xi)f′(xi)thisistheiterativeformto Newton – Rapson.   Newton-Raphson
Newton-Raphson To apreciatedbetterwegonnasolveranexample. fx=e−x−ln(x)  , xo=1 tol= 0.01  
REFERENCES AND BIBLIOGRAPHY ,[object Object]

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