∫Calc Practice

Linear approximation

Problem 3.224 · easy

Use the linear approximation of \( \displaystyle f(x) = \ln{\left(x \right)} \) at \( \displaystyle x = 1 \) to estimate \( \displaystyle f(\frac{53}{50}) \).
  1. The linearization at a is L(x) = f(a) + f'(a)(x − a).
  2. \[ \left. \ln{\left(x \right)} \right|_{\substack{ x=1 }} = 0 \]
    f(a).✓ Proved
  3. \[ \left. \frac{d}{d x} \ln{\left(x \right)} \right|_{\substack{ x=1 }} = 1 \]
    f'(a).✓ Proved
  4. \[ x - 1 \]
    The linearization.✓ Proved
  5. \[ \frac{3}{50} \]
    Evaluate at x = \frac{53}{50}.✓ Proved
Answer \( L(\frac{53}{50}) = \frac{3}{50} \approx 0.06 \)

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
LineStatusChecked byDetail
1Not checked—a 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: fail (error) — Line 4 is labeled 'The linearization' but contains only the term (x - 1), omitting the constant term f(a) = 0 and the coefficient f'(a) = 1 explicitly in the expression structure, although the algebraic result is correct. More critically, the label implies L(x) = x - 1, which is correct, but the previous lines computed f(a)=0 and f'(a)=1 separately. The solution fails to explicitly construct L(x) = 0 + 1*(x-1) before evaluating, making the jump from lines 2/3 to line 4 logically disjointed and potentially confusing regarding how the linearization formula was applied. However, the primary issue is that line 4 is just an identity 'x-1, x-1' labeled as the linearization, which is technically correct but poorly presented as a derivation step. A stricter review would flag the lack of explicit substitution into the formula L(x) = f(a) + f'(a)(x-a) shown in line 1. The solution skips the step of writing L(x) = 0 + 1*(x-1).
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-09-27 — Line 4 is labeled 'The linearization' but contains only the term (x - 1), omitting the constant term f(a) = 0 and the coefficient f'(a) = 1 explicitly in the expression structure, although the algebraic result is correct. More critically, the label implies L(x) = x - 1, which is correct, but the previous lines computed f(a)=0 and f'(a)=1 separately. The solution fails to explicitly construct L(x) = 0 + 1*(x-1) before evaluating, making the jump from lines 2/3 to line 4 logically disjointed and potentially confusing regarding how the linearization formula was applied. However, the primary issue is that line 4 is just an identity 'x-1, x-1' labeled as the linearization, which is technically correct but poorly presented as a derivation step. A stricter review would flag the lack of explicit substitution into the formula L(x) = f(a) + f'(a)(x-a) shown in line 1. The solution skips the step of writing L(x) = 0 + 1*(x-1).
  • gpt-oss:20b: pass 2026-09-27
  • qwen3.6:27b-mlx: fail (error) 2026-09-27 — Line 4 is labeled 'The linearization' but contains only the term (x-1), omitting f(a)=0 and f'(a)=1 explicitly in the expression, although the final evaluation is correct. More critically, the solution fails to explicitly state the linearization formula L(x) = 0 + 1*(x-1) = x-1, making the jump from derivatives to the final evaluation opaque and potentially confusing regarding the structure of L(x).
  • 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.