Intermediate Value Theorem: locating a root
Problem 1.614 · medium
Show that \( \displaystyle x^{3} - 4 = 0 \) 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.
- \[ \left. x^{3} - 4 \right|_{\substack{ x=1 }} = -3 \]f(1).✓ Proved
- \[ \left. x^{3} - 4 \right|_{\substack{ x=2 }} = 4 \]f(2): the opposite sign, so a root lies between.✓ Proved
- \[ \left. x^{3} - 4 \right|_{\substack{ x=\frac{3}{2} }} = - \frac{5}{8} \]f(3/2) ≈ -0.6250.✓ Proved
- \[ \left. x^{3} - 4 \right|_{\substack{ x=\frac{7}{4} }} = \frac{87}{64} \]f(7/4) ≈ 1.3594.✓ Proved
- The sign changes on [3/2, 7/4], so a root lies there.
Answer \( \left[\frac{3}{2}, \frac{7}{4}\right] \)
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 never performs the two required bisection steps; it merely applies the IVT again to the interval [3/2,7/4] and claims that as the final result. This does not demonstrate the requested bisection procedure and therefore is incorrect.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 specified interval.
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 to isolate the root in the specified interval.gpt-oss:20b: fail (error) 2026-10-17 — The solution never performs the two required bisection steps; it merely applies the IVT again to the interval [3/2,7/4] and claims that as the final result. This does not demonstrate the requested bisection procedure and therefore is incorrect.qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the specified interval.gpt-oss:20b: fail (error) 2026-10-14 — The solution does not perform the required two bisection steps; it merely applies the IVT again to the interval [3/2,7/4] and claims a root lies there. The task was to find an interval of length 1/4 containing the root after two bisection iterations, which is not demonstrated.qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to isolate the root in the interval [3/2, 7/4], which has the required length of 1/4.gpt-oss:20b: fail (misleading) 2026-10-14 — The solution skips the explicit bisection steps; it only evaluates f at 3/2 and 7/4, so a student would not learn how the two bisection iterations lead to the interval [3/2,7/4].
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.