Great work deriving the Taylor polynomial and calculating the approximation—let's break down the full process, error estimation, and interval bounds clearly:

1. Verify the Second-Order Taylor Polynomial #

For f(x) = x^(1/3) centered at a=8, the second-order Taylor polynomial follows the form:

P₂(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)²

Calculating each term:

• f(8) = 8^(1/3) = 2
• f'(x) = (1/3)x^(-2/3) → f'(8) = 1/(3*4) = 1/12
• f''(x) = (-2/9)x^(-5/3) → f''(8) = -2/(9*32), so dividing by 2! gives -1/(9*32)

Your polynomial is correct:
P₂(x) = 2 + (1/12)(x-8) - (1/(9*32))(x-8)²

2. Calculate the Approximation for $9^{1/3}$ #

Substitute x=9 into P₂(x):

• (x-8) = 1, so all powers of (x-8) equal 1
• Compute step-by-step:
  • 2 + 1/12 - 1/(9*32)
  • = 2 + 24/288 - 1/288
  • = 2 + 23/288 ≈ 2.07986

This matches your result perfectly.

3. Estimate the Approximation Error #

We use the Lagrange Remainder Formula for error:

|R₂(x)| = |f'''(c)/(3!) * (x-a)³| where c is between 8 and 9

First, find the third derivative:
f'''(x) = (10/27)x^(-8/3) (positive for all x>0, so the error is positive—meaning the exact value is larger than our approximation)

Since c lies in [8,9], c^(-8/3) is a decreasing function, so its maximum value occurs at c=8:
8^(-8/3) = (2^3)^(-8/3) = 2^(-8) = 1/256

Plug into the error formula:
|R₂(9)| < (10/(27*6)) * 1 * (1/256) = 5/(81*256) ≈ 0.000241

4. Tightest Interval Containing $9^{1/3}$ #

Since the error is positive, the exact value satisfies:
P₂(9) < 9^(1/3) < P₂(9) + max_error

Substituting the values:

• Lower bound: ≈2.07986
• Upper bound: ≈2.07986 + 0.000241 ≈2.08010

So the smallest interval containing the exact value is approximately (2.07986, 2.08010).

内容的提问来源于stack exchange，提问作者gbgult