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Linear approximation

Problem 3.347 · easy

Use the linear approximation of \( \displaystyle f(x) = \frac{1}{x} \) at \( \displaystyle x = 10 \) to estimate \( \displaystyle f(\frac{997}{100}) \).

The linearization at a is L(x) = f(a) + f'(a)(x − a).
\[ \left. \frac{1}{x} \right|_{\substack{ x=10 }} = \frac{1}{10} \]
f(a).✓ Proved
\[ \left. \frac{d}{d x} \frac{1}{x} \right|_{\substack{ x=10 }} = - \frac{1}{100} \]
f'(a).✓ Proved
\[ - \frac{x}{100} + 1 \cdot \frac{1}{5} = \frac{1}{5} - \frac{x}{100} \]
The linearization.✓ Proved
\[ \frac{1003}{10000} \]
Evaluate at x = \frac{997}{100}.✓ Proved

Answer \( L(\frac{997}{100}) = \frac{1003}{10000} \approx 0.1003 \)

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: pass
qwen3.6:27b-mlx: fail (error) — The linearization formula in line 4 is incorrect; it uses f(a)=1/5 instead of the correct f(10)=1/10 derived in line 2. Consequently, the final estimate is wrong.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-03 — The linearization formula in line 4 is incorrect; it uses f(a)=1/5 instead of the correct f(10)=1/10 derived in line 2. Consequently, the final estimate is wrong.
gpt-oss:20b: pass 2026-10-03
qwen3.6:27b-mlx: fail (style) 2026-10-03 — [domain objection, downgraded to style] The linearization formula in step 4 is incorrect; it uses f(10)=1/10 and f'(10)=-1/100 but writes 1/5 - x/100. The correct linearization is L(x) = 1/10 - (1/100)(x-10) = 1/5 - x/100 is actually algebraically correct for the expression, but the setup implies L(x) = f(a) + f'(a)(x-a). Let's re-verify: L(x) = 1/10 - 1/100(x-10) = 1/10 - x/100 + 10/100 = 1/10 + 1/10 - x/100 = 2/10 - x/100 = 1/5 - x/100. So the equation in step 4 is actually correct. However, step 5 evaluates this at x = 997/100 = 9.97. L(9.97) = 1/5 - 9.97/100 = 0.2 - 0.0997 = 0.1003 = 1003/10000. The math is correct. Wait, let me look closer at the prompt's constraints. 'Each line... is either an EQUATION... or a SENTENCE'. Step 4 is marked [equation, unchecked]. Step 5 is marked [equation, unchecked]. The prompt says 'do not re-check the algebra of a line marked proved'. But these are marked 'unchecked'. Does 'unchecked' mean I should check them? The prompt says 'a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. It does not explicitly say 'unchecked' lines are exempt from checking. Usually, in these tasks, 'unchecked' implies the CAS didn't verify it, or it's a derived step. But the instructions say 'Check that... each conclusion follows from the lines before it'. Step 4 is a definition/derivation. Step 5 is an evaluation. The derivation in step 4 is correct. The evaluation in step 5 is correct. Is there a subtle error? f(x) = 1/x. a=10. f(10)=1/10. f'(x)=-1/x^2. f'(10)=-1/100. L(x) = 1/10 - 1/100(x-10). L(x) = 1/10 - x/100 + 10/100 = 11/100 - x/100? No. 1/10 + 10/100 = 10/100 + 10/100 = 20/100 = 1/5. So L(x) = 1/5 - x/100. This is correct. Evaluation: x = 997/100. L(997/100) = 1/5 - (997/100)/100 = 1/5 - 997/10000 = 2000/10000 - 997/10000 = 1003/10000. This is correct. Why would it fail? Maybe the sentence in step 1 is too generic? No. Maybe the issue is that step 4 is labeled 'The linearization' but it's an equation `1/5 - x/100 = 1/5 - x/100`. This is a tautology, not a definition. It doesn't explicitly state `L(x) = ...`. It just states an identity. This is a 'style' or 'misleading' issue because it doesn't clearly define L(x). However, step 5 evaluates 'it'. If step 4 doesn't define L(x), step 5 is ambiguous. But usually, these adversarial reviews look for math errors. Let me re-read carefully. 'Subs(1/x, x, 10) = 1/10'. 'Subs(Derivative(1/x, x), x, 10) = -1/100'. Step 4: `1/5 - x/100 = 1/5 - x/100`. This is technically a true statement, but it fails to link the previous steps to the linearization function L(x). It just asserts an identity. A student might not know that `1/5 - x/100` *is* the linearization derived from steps 2 and 3. It skips the logical connection `L(x) = f(a) + f'(a)(x-a) => L(x) = 1/10 - 1/100(x-10) => L(x) = 1/5 - x/100`. By presenting just the simplified form as an identity, it obscures the derivation. Is this an error? It's misleading. But is it a 'fail'? The prompt asks to find what is wrong. If the math is right, is it 'pass'? 'Say pass with empty notes if it is sound.' Is it sound? The logic is sound, but the presentation of step 4 is poor. However, often 'unchecked' equations in these datasets are just intermediate calculations. Let's look for a harder error. Did I calculate 1/5 - 997/10000 correctly? 2000 - 997 = 1003. Yes. Is the approximation direction correct? f(x)=1/x is convex for x>0. Linear approximation is an overestimate? f''(x) = 2/x^3 > 0. So L(x) >= f(x). f(9.97) = 100/997 approx 0.1003009. L(9.97) = 0.1003. 0.1003 < 0.1003009. Wait. If f is convex, the tangent line is below the curve? No, for convex (concave up), the tangent line is below the curve. So L(x) <= f(x). 0.1003 <= 0.1003009. This holds. The estimate is valid. I don't see a mathematical error. I will mark it as pass.
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.