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Linear approximation

Problem 3.113 · easy

Use the linear approximation of \( \displaystyle f(x) = \frac{1}{x} \) at \( \displaystyle x = 10 \) to estimate \( \displaystyle f(\frac{999}{100}) \).

The linearization at a is L(x) = f(a) + f'(a)(x − a).
\[ \left. 1 \frac{1}{x} \right|_{\substack{ x=10 }} = \frac{1}{10} \]
f(a).✓ Proved
\[ \left. \frac{d}{d x} 1 \frac{1}{x} \right|_{\substack{ x=10 }} = - \frac{1}{100} \]
f'(a).✓ Proved
\[ - \frac{x}{100} + 1 \cdot \frac{1}{5} = \frac{1}{5} - \frac{x}{100} \]
The linearization.
\[ \frac{1001}{10000} \]
Evaluate at x = \frac{999}{100}.

Answer \( L(\frac{999}{100}) = \frac{1001}{10000} \approx 0.1001 \)

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: pass
qwen3.6:27b-mlx: fail (error) — The linearization formula in line 4 is incorrect. Using f(10)=1/10 and f'(10)=-1/100, the correct linearization is L(x) = 1/10 - (1/100)(x-10) = 1/5 - x/100. Wait, 1/10 + 10/100 = 1/10 + 1/10 = 1/5. So 1/5 - x/100 is actually correct algebraically. Let me re-read carefully. L(x) = 1/10 - 1/100(x-10) = 1/10 - x/100 + 10/100 = 1/10 - x/100 + 1/10 = 2/10 - x/100 = 1/5 - x/100. The algebra is correct. The evaluation: L(999/100) = 1/5 - (999/100)/100 = 1/5 - 999/10000 = 2000/10000 - 999/10000 = 1001/10000. The solution is actually correct. Why did I think it was wrong? Let me double check the problem statement. f(x)=1/x at x=10. Estimate f(999/100). 999/100 = 9.99. This is close to 10. The calculation holds. The sentences are standard definitions. There is no error.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The linearization formula in line 4 is incorrect. Using f(10)=1/10 and f'(10)=-1/100, the correct linearization is L(x) = 1/10 - (1/100)(x-10) = 1/5 - x/100. Wait, 1/10 + 10/100 = 1/10 + 1/10 = 1/5. So 1/5 - x/100 is actually correct algebraically. Let me re-read carefully. L(x) = 1/10 - 1/100(x-10) = 1/10 - x/100 + 10/100 = 1/10 - x/100 + 1/10 = 2/10 - x/100 = 1/5 - x/100. The algebra is correct. The evaluation: L(999/100) = 1/5 - (999/100)/100 = 1/5 - 999/10000 = 2000/10000 - 999/10000 = 1001/10000. The solution is actually correct. Why did I think it was wrong? Let me double check the problem statement. f(x)=1/x at x=10. Estimate f(999/100). 999/100 = 9.99. This is close to 10. The calculation holds. The sentences are standard definitions. There is no error.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the linear approximation formula, computes the derivative and function value at the center, and evaluates the linearization at the target point. The arithmetic is correct.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The linearization formula in step 4 is incorrect; it uses 1/5 instead of the correct f(a) = 1/10. This leads to an incorrect final estimate.
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.