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Intermediate Value Theorem: locating a root

Problem 1.683 · medium

Show that \( \displaystyle x^{3} - 2 x - 3 = 0 \) has a solution in \( \displaystyle [1, 2] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.

f is continuous on [1, 2] (it is built from continuous functions), so the IVT applies.
Reviewed
\[ \left. x^{3} - 2 x - 3 \right|_{\substack{ x=1 }} = -4 \]
f(1).✓ Proved
\[ \left. x^{3} - 2 x - 3 \right|_{\substack{ x=2 }} = 1 \]
f(2): the opposite sign, so a root lies between.✓ Proved
\[ \left. x^{3} - 2 x - 3 \right|_{\substack{ x=\frac{3}{2} }} = - \frac{21}{8} \]
f(3/2) ≈ -2.6250.✓ Proved
\[ \left. x^{3} - 2 x - 3 \right|_{\substack{ x=\frac{7}{4} }} = - \frac{73}{64} \]
f(7/4) ≈ -1.1406.✓ Proved
The sign changes on [7/4, 2], so a root lies there.
Reviewed

Answer \( \left[\frac{7}{4}, 2\right] \)

Lines: 4 proved, 2 reviewed. 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	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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	mpmath's bisection finds the root inside the claimed interval

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in an interval of length 1/4.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in an interval of length 1/4.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution performs two bisection steps but fails to explicitly state the intermediate conclusion that the root lies in [3/2, 2] after the first step. It jumps from evaluating f(3/2) to evaluating f(7/4) and concluding the root is in [7/4, 2], skipping the logical step of narrowing the interval to [3/2, 2] first, which is required to justify why 7/4 is the correct midpoint to evaluate next.
gpt-oss:20b: fail (misleading) 2026-10-16 — The solution skips the explicit statement that the sign change at x=3/2 shows the root lies in [3/2,2] before proceeding to the second bisection. A student would miss the logical link between the computed values and the bisection steps.
qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer returned a non-object
gpt-oss:20b: fail (error) 2026-10-15 — The solution stops after one bisection step; it does not perform the second step required to obtain an interval of length 1/4. The final interval [7/4,2] has length 1/4, but the reasoning does not justify that this is the result of two bisection steps.

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/ivt_root, checked 2026-10-18 with SymPy 1.14.0.