∫Calc Practice

Linear approximation

Problem 3.225 · easy

Use the linear approximation of \( \displaystyle f(x) = \sqrt{x} \) at \( \displaystyle x = 81 \) to estimate \( \displaystyle f(\frac{408}{5}) \).
  1. The linearization at a is L(x) = f(a) + f'(a)(x − a).
    Reviewed
  2. \[ \left. \sqrt{x} \right|_{\substack{ x=81 }} = 9 \]
    f(a).✓ Proved
  3. \[ \left. \frac{d}{d x} \sqrt{x} \right|_{\substack{ x=81 }} = \frac{1}{18} \]
    f'(a).✓ Proved
  4. \[ \frac{x}{18} + \frac{9}{2} \]
    The linearization.✓ Proved
  5. \[ \frac{271}{30} \]
    Evaluate at x = \frac{408}{5}.✓ Proved
Answer \( L(\frac{408}{5}) = \frac{271}{30} \approx 9.03333 \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt
LineStatusChecked byDetail
1Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the estimate is within Taylor's error bound of the true value

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly applies the linearization formula, computes the derivative and function value at the point of tangency, and evaluates the linear approximation at the specified point. The arithmetic is correct.
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the linearization formula, computes the derivative and function value at the point of tangency, and evaluates the linear approximation at the specified point. The arithmetic is correct.
  • gpt-oss:20b: pass 2026-09-27
  • qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the linearization formula, computes the function value and derivative at the point of tangency, constructs the linear approximation, and evaluates it at the specified point. All steps are mathematically sound.
  • gpt-oss:20b: pass 2026-09-27

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/linear_approximation, checked 2026-09-27 with SymPy 1.14.0.