NAG Library Manual, Mark 30.1
Interfaces:  FL   CL   CPP   AD 

NAG FL Interface Introduction
Example description

 C05AZF Example Program Results

  Iterations

  X = 0.00000   FX =  0.1000E+01   IND = 2
  X = 1.00000   FX = -0.6321E+00   IND = 3
  X = 0.61270   FX = -0.7081E-01   IND = 4
  X = 0.56707   FX =  0.1154E-03   IND = 4
  X = 0.56714   FX = -0.9448E-06   IND = 4
  X = 0.56713   FX =  0.1473E-04   IND = 4
  X = 0.56714   FX = -0.9448E-06   IND = 4

  Solution

  X = 0.56714   Y = 0.56713