Linear approximation
Problem 3.117 · easy
Use the linear approximation of \( \displaystyle f(x) = \frac{1}{x} \) at \( \displaystyle x = 4 \) to estimate \( \displaystyle f(\frac{403}{100}) \).
- The linearization at a is L(x) = f(a) + f'(a)(x − a).
- \[ \left. 1 \frac{1}{x} \right|_{\substack{ x=4 }} = \frac{1}{4} \]f(a).✓ Proved
- \[ \left. \frac{d}{d x} 1 \frac{1}{x} \right|_{\substack{ x=4 }} = - \frac{1}{16} \]f'(a).✓ Proved
- \[ - \frac{x}{16} + 1 \cdot \frac{1}{2} = \frac{1}{2} - \frac{x}{16} \]The linearization.
- \[ \frac{397}{1600} \]Evaluate at x = \frac{403}{100}.
Answer \( L(\frac{403}{100}) = \frac{397}{1600} \approx 0.248125 \)
Lines: 2 proved, 3 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. How we verify
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 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 5 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 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 line 4 is incorrect. It should be L(x) = 1/4 - 1/16(x - 4), which simplifies to 1/2 - x/16. Wait, 1/4 - 1/16(x-4) = 1/4 - x/16 + 4/16 = 1/4 - x/16 + 1/4 = 1/2 - x/16. The equation in line 4 is actually correct algebraically. Let me re-read carefully. Line 4 says '1/2 - x/16'. This is the correct simplified form of the linearization. Line 5 evaluates this at x = 403/100. 1/2 - (403/100)/16 = 1/2 - 403/1600 = 800/1600 - 403/1600 = 397/1600. The math is correct. Why did I think it was wrong? Let me check the prompt again. 'The linearization at a is L(x) = f(a) + f'(a)(x - a).' This is correct. f(4) = 1/4. f'(4) = -1/16. L(x) = 1/4 - 1/16(x-4) = 1/4 - x/16 + 1/4 = 1/2 - x/16. This is correct. Evaluation: 1/2 - 403/1600 = 397/1600. This is correct. Is there a subtle error? The problem asks for the linear approximation. The solution provides it. The steps are logically sound. The algebra is verified by the 'proved' tag. The sentences are standard definitions. There is no error. I will mark it as pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The linearization formula in line 4 is incorrect. It should be L(x) = 1/4 - 1/16(x - 4), which simplifies to 1/2 - x/16. Wait, 1/4 - 1/16(x-4) = 1/4 - x/16 + 4/16 = 1/4 - x/16 + 1/4 = 1/2 - x/16. The equation in line 4 is actually correct algebraically. Let me re-read carefully. Line 4 says '1/2 - x/16'. This is the correct simplified form of the linearization. Line 5 evaluates this at x = 403/100. 1/2 - (403/100)/16 = 1/2 - 403/1600 = 800/1600 - 403/1600 = 397/1600. The math is correct. Why did I think it was wrong? Let me check the prompt again. 'The linearization at a is L(x) = f(a) + f'(a)(x - a).' This is correct. f(4) = 1/4. f'(4) = -1/16. L(x) = 1/4 - 1/16(x-4) = 1/4 - x/16 + 1/4 = 1/2 - x/16. This is correct. Evaluation: 1/2 - 403/1600 = 397/1600. This is correct. Is there a subtle error? The problem asks for the linear approximation. The solution provides it. The steps are logically sound. The algebra is verified by the 'proved' tag. The sentences are standard definitions. There is no error. I will mark it as pass.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-27 — The linearization formula in line 4 is incorrect; it should be 1/4 - 1/16(x-4), not 1/2 - x/16. Although the final numerical answer is correct due to a coincidence of errors, the derived function is wrong.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: fail (error) 2026-09-26 — The linearization formula in step 4 is incorrect. It should be L(x) = 1/4 - 1/16(x - 4), which simplifies to 1/2 - x/16 only if the constant term is handled correctly, but the expression '1/2 - x/16' implies L(x) = 1/2 - x/16. Let's check: L(x) = 1/4 - (x-4)/16 = 4/16 - x/16 + 4/16 = 8/16 - x/16 = 1/2 - x/16. The algebra is actually correct. Wait, let me re-read carefully. Step 4 says '1/2 - x/16, 1/2 - x/16 -- The linearization.' This is the simplified form of 1/4 - 1/16(x-4). 1/4 + 4/16 = 1/4 + 1/4 = 1/2. So L(x) = 1/2 - x/16 is correct. Step 5 evaluates this at x = 403/100. L(403/100) = 1/2 - (403/100)/16 = 1/2 - 403/1600 = 800/1600 - 403/1600 = 397/1600. The solution is mathematically correct. Why would it fail? Let me check the prompt again. 'Use the linear approximation... to estimate'. The steps are correct. Is there a subtle error? Step 4 presents the linearization as 1/2 - x/16. This is a valid simplification. Step 5 evaluates it. The result matches the stated answer. The sentences are: 1. Definition. 2. f(a). 3. f'(a). 4. Linearization. 5. Evaluation. All seem correct. Let me look for 'misleading' or 'style'. The sentence in step 4 just says 'The linearization.' It doesn't explicitly show the substitution into the formula from step 1, but step 4 is marked as an equation proved by CAS. The sentence is just a label. Is it misleading? No. Is it an error? No. Wait, look at step 4 again. `1/2 - x/16`. This is `L(x)`. Step 5 `397/1600`. This is `L(403/100)`. Everything is correct. I will mark it as pass.gpt-oss:20b: pass 2026-09-26
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 2026-10-13 with SymPy 1.14.0.