Intermediate Value Theorem: locating a root
Problem 1.825 · medium
Show that \( \displaystyle x^{3} - x - 1 = 0 \) 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
- \[ \left. x^{3} - x - 1 \right|_{\substack{ x=1 }} = -1 \]f(1).✓ Proved
- \[ \left. x^{3} - x - 1 \right|_{\substack{ x=2 }} = 5 \]f(2): 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 [5/4, 3/2], so a root lies there.Reviewed
Answer \( \left[\frac{5}{4}, \frac{3}{2}\right] \)
✓ Nihil obstat Lines: 4 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 | ✓ 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: passqwen3.6:27b-mlx: pass — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the specified interval.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-11-12 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the specified interval.gpt-oss:20b: pass 2026-11-12qwen3.6:27b-mlx: pass 2026-11-11 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the interval [5/4, 3/2], which has the required length of 1/4.gpt-oss:20b: pass 2026-11-11qwen3.6:27b-mlx: pass 2026-11-11 — 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 2026-11-11
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 2026-11-12 with SymPy 1.14.0.