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  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. double f(double x) {
  5.   return x - 0.2 * (pow(x, 2) + 6);
  6. }
  7.  
  8. double solve(double x0, double eps) {
  9.   double x = x0;
  10.   int iter = 0;
  11.  
  12.   while (fabs(0.2 * (pow(x, 2) + 6) - x) > eps) {
  13.     x = 0.2 * (pow(x, 2) + 6);
  14.     iter++;
  15.   }
  16.  
  17.   printf("Итераций: %d\n", iter);
  18.   return x;
  19. }
  20.  
  21. int main() {
  22.   double x0 = 0;
  23.   double eps =1e-20;
  24.   double root = solve(x0, eps);
  25.  
  26.   printf("Решение: x = %lf\n", root);
  27.   printf("Значение f(x) = %lf\n", f(root));
  28.  
  29.   return 0;
  30. }
  31.  
Success #stdin #stdout 0s 5284KB
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Standard input is empty
stdout
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Итераций: 150
Решение: x = 2.000000
Значение f(x) = 0.000000
https://ideone.com/C8Q3Fb
language:
C (gcc 8.3)
created:
4 days ago
visibility:
public

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