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Tangent lines

Problem 2.738 · medium Video

Find an equation of the tangent line to \( \displaystyle y = - x + 3 \sin{\left(x \right)} \) at \( \displaystyle x = \pi \).

The tangent line passes through the point (a, f(a)) and has slope f'(a).
Reviewed
\[ \left. - x + 3 \sin{\left(x \right)} \right|_{\substack{ x=\pi }} = - \pi \]
The point of tangency.✓ Proved
\[ \frac{d}{d x} \left(- x + 3 \sin{\left(x \right)}\right) = 3 \cos{\left(x \right)} - 1 \]
Differentiate.✓ Proved
\[ \left. 3 \cos{\left(x \right)} - 1 \right|_{\substack{ x=\pi }} = -4 \]
The slope at the point.✓ Proved
\[ - 4 x + 3 \pi \]
Point-slope form, then simplify.Reviewed

Answer \( y = - 4 x + 3 \pi \)

The graph of y = -x + 3sin(x) near x = pi. A secant line through the point (pi, -pi) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -4, giving y = -4x + 3pi.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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 line meets the curve at x = a, and its slope matches a central difference quotient of f there

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the point of tangency and the slope, and derives the correct linear equation. The steps are logically sound and algebraically correct.

Senior review claude-sonnet-5-5, 2026-10-12: pass — The math checks out: f(π) = -π, f'(x) = 3cos x - 1, f'(π) = -4, and y + π = -4(x - π) gives y = -4x + 3π. Line 5 is a trivial identity that bundles point-slope form with simplification, but it is correct and the contract allows that.

Every verdict on record (7)

claude-sonnet-5-5: pass 2026-10-12 — The math checks out: f(π) = -π, f'(x) = 3cos x - 1, f'(π) = -4, and y + π = -4(x - π) gives y = -4x + 3π. Line 5 is a trivial identity that bundles point-slope form with simplification, but it is correct and the contract allows that.
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the point of tangency and the slope, and derives the correct linear equation. The steps are logically sound and algebraically correct.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The y-coordinate of the point of tangency is calculated incorrectly as -pi instead of pi. Consequently, the final equation is wrong.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the point of tangency and the slope, and the final equation is algebraically correct.
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/tangent_line, checked 2026-10-13 with SymPy 1.14.0.