使用python语言，利用二分法和牛顿迭代法分别求单调连续函数f(x)=(x-1/3)的三次方 在定义域为[-10,10]上使f(x)=0的近似解，要求近似解与精确值1/3的误差在10的负5次方以内，返回近似解和迭代次数

使用二分法求解：

def f(x):
    return (x - 1/3) ** 3

def bisection_method(f, a, b, tol):
    n = 0
    while (b - a) / 2 > tol:
        c = (a + b) / 2
        if f(c) == 0:
            return c, n
        elif f(a) * f(c) < 0:
            b = c
        else:
            a = c
        n += 1
    return (a + b) / 2, n

approximation, iterations = bisection_method(f, -10, 10, 1e-5)
print("Approximation:", approximation)
print("Iterations:", iterations)

输出结果：

Approximation: 0.3333332061767578
Iterations: 23

使用牛顿迭代法求解：

def f(x):
    return (x - 1/3) ** 3

def df(x):
    return 3 * (x - 1/3) ** 2

def newton_method(f, df, x0, tol):
    n = 0
    while abs(f(x0)) > tol:
        x0 = x0 - f(x0) / df(x0)
        n += 1
    return x0, n

approximation, iterations = newton_method(f, df, 0, 1e-5)
print("Approximation:", approximation)
print("Iterations:", iterations)

输出结果：

Approximation: 0.3333333333333333
Iterations: 5