Home›Calculus 1›Intermediate Value Theorem: locating a root›Problem 1.744

Intermediate Value Theorem: locating a root

Problem 1.744 · medium

Show that \( \displaystyle x^{3} - 3 x - 5 = 0 \) has a solution in \( \displaystyle [2, 3] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.

f is continuous on [2, 3] (it is built from continuous functions), so the IVT applies.
Reviewed
\[ \left. x^{3} - 3 x - 5 \right|_{\substack{ x=2 }} = -3 \]
f(2).✓ Proved
\[ \left. x^{3} - 3 x - 5 \right|_{\substack{ x=3 }} = 13 \]
f(3): the opposite sign, so a root lies between.✓ Proved
\[ \left. x^{3} - 3 x - 5 \right|_{\substack{ x=\frac{5}{2} }} = \frac{25}{8} \]
f(5/2) ≈ 3.1250.✓ Proved
\[ \left. x^{3} - 3 x - 5 \right|_{\substack{ x=\frac{9}{4} }} = - \frac{23}{64} \]
f(9/4) ≈ -0.3594.✓ Proved
The sign changes on [9/4, 5/2], so a root lies there.
Reviewed

Answer \( \left[\frac{9}{4}, \frac{5}{2}\right] \)

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 the interval [9/4, 5/2], which has the required length of 1/4.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-21 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the interval [9/4, 5/2], which has the required length of 1/4.
gpt-oss:20b: pass 2026-10-21
qwen3.6:27b-mlx: pass 2026-10-20 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the interval [9/4, 5/2], which has the required length of 1/4.
gpt-oss:20b: fail (misleading) 2026-10-20 — The solution skips the explicit bisection steps; it only evaluates the function at 5/2 and 9/4 and then states a sign change on [9/4,5/2] without showing that this interval results from two successive bisections starting from [2,3].
qwen3.6:27b-mlx: pass 2026-10-20 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the interval [9/4, 5/2], which has the required length of 1/4.
gpt-oss:20b: pass 2026-10-20

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 2026-10-21 with SymPy 1.14.0.