Linear approximation
Problem 3.344 · easy
Use the linear approximation of \( \displaystyle f(x) = \sqrt{x} \) at \( \displaystyle x = 36 \) to estimate \( \displaystyle f(\frac{181}{5}) \).
- The linearization at a is L(x) = f(a) + f'(a)(x − a).
- \[ \left. \sqrt{x} \right|_{\substack{ x=36 }} = 6 \]f(a).✓ Proved
- \[ \left. \frac{d}{d x} \sqrt{x} \right|_{\substack{ x=36 }} = \frac{1}{12} \]f'(a).✓ Proved
- \[ \frac{x}{12} + 3 \]The linearization.✓ Proved
- \[ \frac{361}{60} \]Evaluate at x = \frac{181}{5}.✓ Proved
Answer \( L(\frac{181}{5}) = \frac{361}{60} \approx 6.01667 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the estimate is within Taylor's error bound of the true value |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The linearization formula in step 4 is incorrect; it uses 3 instead of the correct constant term 5.5 (or 11/2). The correct linearization is L(x) = 6 + (1/12)(x - 36) = x/12 + 5.5. Consequently, the final evaluation is wrong.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-03 — The linearization formula in step 4 is incorrect; it uses 3 instead of the correct constant term 5.5 (or 11/2). The correct linearization is L(x) = 6 + (1/12)(x - 36) = x/12 + 5.5. Consequently, the final evaluation is wrong.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly applies the linearization formula, computes the derivative and function value at the specified point, and evaluates the resulting linear function at the target input.gpt-oss:20b: pass 2026-10-03
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 2026-10-03 with SymPy 1.14.0.