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Intermediate Value Theorem: locating a root

Problem 1.642 · medium

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

f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
Reviewed
\[ \left. x^{3} - x + 1 \right|_{\substack{ x=-2 }} = -5 \]
f(-2).✓ Proved
\[ \left. x^{3} - x + 1 \right|_{\substack{ x=-1 }} = 1 \]
f(-1): the opposite sign, so a root lies between.✓ Proved
\[ \left. x^{3} - x + 1 \right|_{\substack{ x=- \frac{3}{2} }} = - \frac{7}{8} \]
f(-3/2) ≈ -0.8750.✓ Proved
\[ \left. x^{3} - x + 1 \right|_{\substack{ x=- \frac{5}{4} }} = \frac{19}{64} \]
f(-5/4) ≈ 0.2969.✓ Proved
The sign changes on [-3/2, -5/4], so a root lies there.
Reviewed

Answer \( \left[- \frac{3}{2}, - \frac{5}{4}\right] \)

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. The logic and calculations are sound.

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. The logic and calculations are sound.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to identify the correct interval containing the root.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The final interval [-3/2, -5/4] is invalid because the lower bound (-1.5) is greater than the upper bound (-1.25). The correct interval containing the root is [-5/4, -3/2].
gpt-oss:20b: pass 2026-10-14

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.