We had to approximate the root of x^3+3x-2 for calc1 on an interval of size 1/8, I wrote this program to find it on an interval of size 1/(2^10000). It only takes a few minutes to run.

bisec.py

def gcd(x: int, y: int):
  while(y):
    x, y = y, x % y
  return x

def frac(a: int, b: int):
  return '\\seqsplit{' + str(a//gcd(a, b)) + '\\div' + str(b//gcd(a,b)) + '}'

def func(a, n):
  """ x=a/2**n, f(x) = x^3+3x-2 """
  p1, q1 = a**3, 1 << (3*n) # first coefficient
  p2, q2 = 3 * a, 1 << n    # second coefficient
  den = (q1 * q2)
  num = (p1 * q2 + p2 * q1 - 2 * den) 
  return frac(num, den), (den < 0 or num < 0) and not (den < 0 and num < 0)

# start with 1 / 2**1 = 1/2
a, n = 1, 1

for i in range(10000):
  res, neg = func(a, n)
  print(f'Neem het midden van ${frac(a-1, 1 << n)}$ en ${frac(a+1, 1 << n)}$. Dat is ${frac(a, 1 << n)}$. $f({frac(a, 1 << n)})={res}{"<" if neg else ">"}0$. \par')
  n += 1
  a = a*2 + (1 if neg else -1)