Intermediate Value Theorem: locating a root
Problem 1.641 · medium
Show that \( \displaystyle x^{3} - 2 x - 5 = 0 \) 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.
- \[ \left. x^{3} - 2 x - 5 \right|_{\substack{ x=2 }} = -1 \]f(2).✓ Proved
- \[ \left. x^{3} - 2 x - 5 \right|_{\substack{ x=3 }} = 16 \]f(3): the opposite sign, so a root lies between.✓ Proved
- \[ \left. x^{3} - 2 x - 5 \right|_{\substack{ x=\frac{5}{2} }} = \frac{45}{8} \]f(5/2) ≈ 5.6250.✓ Proved
- \[ \left. x^{3} - 2 x - 5 \right|_{\substack{ x=\frac{9}{4} }} = \frac{121}{64} \]f(9/4) ≈ 1.8906.✓ Proved
- The sign changes on [2, 9/4], so a root lies there.
Answer \( \left[2, \frac{9}{4}\right] \)
Lines: 4 proved, 2 not checked. 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 | Not checked | — | 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 | Not checked | — | 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: inconclusive — reviewer returned a non-objectqwen3.6:27b-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: inconclusive 2026-10-18 — reviewer returned a non-objectqwen3.6:27b-mlx: pass 2026-10-16 — 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 2026-10-16qwen3.6:27b-mlx: pass 2026-10-14 — 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 is sound and the calculations are correct.gpt-oss:20b: fail (misleading) 2026-10-14 — The solution skips the explicit bisection steps; it merely evaluates the function at 5/2 and 9/4 and then states a sign change on [2,9/4] without showing that the bisection algorithm actually produced that interval. A student would learn that one can jump to the final interval without performing the required two bisection iterations.
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.