e^x -3xにおいて区間[0,1]の解を二分法で求める

Code

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define FUNC(x) exp(x) - 3 * x

double a = 0.0;
double b = 1.0;
double c = 0.0;
double eps = 0.00001;
double fa, fb, fc;
double ans;
int i = 0;

int main() {
    int n = log((b - a) / eps) / log(2.0) + 1;
    while (i < n) {
        c = (a + b) / 2.0;
        fa = FUNC(a);
        fc = FUNC(c);

        if (fa * fc < 0) {
            b = c;
        } else {
            a = c;
        }
        if (fabs(b - a) < eps) {
            ans = c;
        }
        printf("i:%d c=%f fc=%f\n", i, c, fc);
        i++;
    }
    printf("---------------------------------\n");
    printf("Ans: c = %f\tf(c) = %f\n", ans, FUNC(ans));
}

Output

i:0 c=0.500000 fc=0.148721
i:1 c=0.750000 fc=-0.133000
i:2 c=0.625000 fc=-0.006754
i:3 c=0.562500 fc=0.067555
i:4 c=0.593750 fc=0.029516
i:5 c=0.609375 fc=0.011156
i:6 c=0.617188 fc=0.002145
i:7 c=0.621094 fc=-0.002319
i:8 c=0.619141 fc=-0.000091
i:9 c=0.618164 fc=0.001026
i:10 c=0.618652 fc=0.000468
i:11 c=0.618896 fc=0.000188
i:12 c=0.619019 fc=0.000049
i:13 c=0.619080 fc=-0.000021
i:14 c=0.619049 fc=0.000014
i:15 c=0.619064 fc=-0.000003
i:16 c=0.619057 fc=0.000005
---------------------------------
Ans: c = 0.619057       f(c) = 0.000005