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Tangent lines

Problem 2.770 · medium Video

Find an equation of the tangent line to \( \displaystyle y = 2 \sqrt{x} - 3 x - 2 \) at \( \displaystyle x = 1 \).

The tangent line passes through the point (a, f(a)) and has slope f'(a).
Reviewed
\[ \left. 2 \sqrt{x} - 3 x - 2 \right|_{\substack{ x=1 }} = -3 \]
The point of tangency.✓ Proved
\[ \frac{d}{d x} \left(2 \sqrt{x} - 3 x - 2\right) = -3 + \frac{1}{\sqrt{x}} \]
Differentiate.✓ Proved
\[ \left. -3 + 1 \frac{1}{\sqrt{x}} \right|_{\substack{ x=1 }} = -2 \]
The slope at the point.✓ Proved
\[ - 2 x - 1 \]
Point-slope form, then simplify.

Answer \( y = - 2 x - 1 \)

The graph of y = 2sqrt(x) - 3x - 2 near x = 1. A secant line through the point (1, -3) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -2, giving y = -2x - 1.

Lines: 3 proved, 1 reviewed, 1 not checked. 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	claude-sonnet-5-5	claude-sonnet-5-5: Line 1 is a sentence stating the standard definition of the tangent line: it passes through (a, f(a)) with slope f'(a). It is true and is not an algebraic equivalence.
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	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 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: fail (error) — The solution calculates the y-coordinate of the point of tangency as -3, but the correct value is f(1) = 2*sqrt(1) - 3*1 - 2 = -3. Wait, 2 - 3 - 2 = -3. The point is (1, -3). The slope is -2. The line is y - (-3) = -2(x - 1) => y + 3 = -2x + 2 => y = -2x - 1. The algebra is actually correct. Let me re-read carefully. f(1) = 2(1) - 3(1) - 2 = 2 - 3 - 2 = -3. Point (1, -3). Slope f'(1) = 1/sqrt(1) - 3 = 1 - 3 = -2. Equation: y - (-3) = -2(x - 1) => y + 3 = -2x + 2 => y = -2x - 1. The stated answer is y = -2x - 1. The solution steps are correct. Why did I think it was wrong? Let me check the subtraction again. 2 - 3 = -1. -1 - 2 = -3. Correct. Derivative: 2*(1/2)*x^(-1/2) - 3 = 1/sqrt(x) - 3. At x=1, 1 - 3 = -2. Correct. Point slope: y - y1 = m(x - x1). y - (-3) = -2(x - 1). y + 3 = -2x + 2. y = -2x - 1. Correct. The solution is sound.

Senior review claude-sonnet-5-5, 2026-10-12: pass — The solution is correct: f(1) = -3, f'(x) = 1/sqrt(x) - 3, f'(1) = -2, and the point-slope form y + 3 = -2(x - 1) gives y = -2x - 1. Line 5 is slightly terse, since it combines point-slope substitution and simplification, but it is labelled as such and acceptable.

qwen3.6:27b-mlx: dismiss — The objection is labeled an error, but its own reasoning confirms f(1) = -3, f'(1) = -2 and the line y = -2x - 1, so it identifies no defect.

Every verdict on record (7)

claude-sonnet-5-5: pass 2026-10-12 — The solution is correct: f(1) = -3, f'(x) = 1/sqrt(x) - 3, f'(1) = -2, and the point-slope form y + 3 = -2(x - 1) gives y = -2x - 1. Line 5 is slightly terse, since it combines point-slope substitution and simplification, but it is labelled as such and acceptable.
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution calculates the y-coordinate of the point of tangency as -3, but the correct value is f(1) = 2*sqrt(1) - 3*1 - 2 = -3. Wait, 2 - 3 - 2 = -3. The point is (1, -3). The slope is -2. The line is y - (-3) = -2(x - 1) => y + 3 = -2x + 2 => y = -2x - 1. The algebra is actually correct. Let me re-read carefully. f(1) = 2(1) - 3(1) - 2 = 2 - 3 - 2 = -3. Point (1, -3). Slope f'(1) = 1/sqrt(1) - 3 = 1 - 3 = -2. Equation: y - (-3) = -2(x - 1) => y + 3 = -2x + 2 => y = -2x - 1. The stated answer is y = -2x - 1. The solution steps are correct. Why did I think it was wrong? Let me check the subtraction again. 2 - 3 = -1. -1 - 2 = -3. Correct. Derivative: 2*(1/2)*x^(-1/2) - 3 = 1/sqrt(x) - 3. At x=1, 1 - 3 = -2. Correct. Point slope: y - y1 = m(x - x1). y - (-3) = -2(x - 1). y + 3 = -2x + 2. y = -2x - 1. Correct. The solution is sound.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the point of tangency and the slope, and the final equation is algebraically correct.
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 derives the correct tangent line equation.
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.