Intermediate Value Theorem: locating a root
Problem 1.616 · medium
Show that \( \displaystyle x^{3} - 9 = 0 \) 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.
- \[ \left. x^{3} - 9 \right|_{\substack{ x=2 }} = -1 \]f(2).✓ Proved
- \[ \left. x^{3} - 9 \right|_{\substack{ x=3 }} = 18 \]f(3): the opposite sign, so a root lies between.✓ Proved
- \[ \left. x^{3} - 9 \right|_{\substack{ x=\frac{5}{2} }} = \frac{53}{8} \]f(5/2) ≈ 6.6250.✓ Proved
- \[ \left. x^{3} - 9 \right|_{\substack{ x=\frac{9}{4} }} = \frac{153}{64} \]f(9/4) ≈ 2.3906.✓ 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: fail (error) — The solution skips the required two bisection steps; it simply notes a sign change at 9/4 and concludes the interval [2,9/4] without demonstrating the bisection process. This misrepresents the method and does not satisfy the problem’s instructions.qwen3.6:27b-mlx: pass — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection. The final interval [2, 9/4] has length 1/4 and contains the root, as f(2) < 0 and f(9/4) > 0.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection. The final interval [2, 9/4] has length 1/4 and contains the root, as f(2) < 0 and f(9/4) > 0.gpt-oss:20b: fail (error) 2026-10-17 — The solution skips the required two bisection steps; it simply notes a sign change at 9/4 and concludes the interval [2,9/4] without demonstrating the bisection process. This misrepresents the method and does not satisfy the problem’s instructions.qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution fails to perform the second step of bisection. It calculates f(5/2) and f(9/4) but does not calculate f(9/4 + 2/4) = f(11/4) or check tgpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution claims the sign changes on [2, 9/4], but f(2) is negative and f(9/4) is positive, which is correct. However, the problem asks for an interval of length 1/4. The interval [2, 9/4] has length 1/4. Wait, let's re-read carefully. f(2) = -1 (negative). f(2.5) = 6.625 (positive). So the root is in [2, 2.5]. Step 2 of bisection: midpoint of [2, 2.5] is 2.25 (9/4). f(2.25) = 2.39 (positive). So the root is in [2, 2.25]. The length is 0.25 = 1/4. The interval is [2, 9/4]. The logic holds. Why did I think it failed? Let me re-check the signs. f(2) = 4-9 = -5? No, 2^3 = 8. 8-9 = -1. Correct. f(3) = 27-9 = 18. Correct. Midpoint 2.5. 2.5^3 = 15.625. 15.625 - 9 = 6.625. Positive. Root in [2, 2.5]. Midpoint 2.25. 2.25^3 = 11.390625. 11.390625 - 9 = 2.390625. Positive. Root in [2, 2.25]. Length 0.25. Interval [2, 9/4]. The solution is correct. Wait, is there a subtle error? 'The sign changes on [2, 9/4]'. f(2) < 0, f(9/4) > 0. Yes, sign changes. The conclusion is correct. The steps are correct. Is there a style issue? 'Subs(...)' is not standard math notation but the prompt says equations are checked by CAS. The sentences are: 1. IVT applies. 6. Sign changes, so root lies there. These are correct. Verdict: pass.gpt-oss:20b: fail (error) 2026-10-14 — The solution skips the required two bisection steps; it merely evaluates f at 9/4 and concludes the interval. This does not demonstrate the bisection method and misleads the reader about the procedure.
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 2026-10-17 with SymPy 1.14.0.