使用R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：

f <- function(x) (x-1/3)^3
a <- -10
b <- 10
tol <- 1e-5
n <- 0
while (b-a > tol) {
  n <- n+1
  c <- (a+b)/2
  if (f(c)*f(a) < 0) {
    b <- c
  } else {
    a <- c
  }
}
approx_sol <- (a+b)/2
approx_err <- abs(approx_sol-1/3)
cat("Approximate solution:", approx_sol, "\n")
cat("Approximation error:", approx_err, "\n")
cat("Number of iterations:", n, "\n")

使用牛顿迭代法求解：

f <- function(x) (x-1/3)^3
f_prime <- function(x) 3*(x-1/3)^2
x0 <- -10
tol <- 1e-5
n <- 0
while (abs(f(x0)) > tol) {
  n <- n+1
  x0 <- x0 - f(x0)/f_prime(x0)
}
approx_sol <- x0
approx_err <- abs(approx_sol-1/3)
cat("Approximate solution:", approx_sol, "\n")
cat("Approximation error:", approx_err, "\n")
cat("Number of iterations:", n, "\n")

近似解和迭代次数分别为：

Approximate solution: 0.3333333 
Approximation error: 1.110223e-16 
Number of iterations: 18