Intermediate Value Theorem: locating a root
Problem 1.643 · medium
Show that \( \displaystyle x^{3} + x - 3 = 0 \) 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.
- \[ \left. x^{3} + x - 3 \right|_{\substack{ x=1 }} = -1 \]f(1).✓ Proved
- \[ \left. x^{3} + x - 3 \right|_{\substack{ x=2 }} = 7 \]f(2): the opposite sign, so a root lies between.✓ Proved
- \[ \left. x^{3} + x - 3 \right|_{\substack{ x=\frac{3}{2} }} = \frac{15}{8} \]f(3/2) ≈ 1.8750.✓ Proved
- \[ \left. x^{3} + x - 3 \right|_{\substack{ x=\frac{5}{4} }} = \frac{13}{64} \]f(5/4) ≈ 0.2031.✓ Proved
- The sign changes on [1, 5/4], so a root lies there.
Answer \( \left[1, \frac{5}{4}\right] \)
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 (misleading) — The solution skips the explicit two bisection steps: it jumps from f(1) and f(2) to f(5/4) without showing that 5/4 is the midpoint of the first subinterval [1,3/2] and that the root lies in [1,5/4] after two bisections. This could mislead a student into thinking the interval was obtained directly rather than by the required two bisection steps.qwen3.6:27b-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: fail (misleading) 2026-10-18 — The solution skips the explicit two bisection steps: it jumps from f(1) and f(2) to f(5/4) without showing that 5/4 is the midpoint of the first subinterval [1,3/2] and that the root lies in [1,5/4] after two bisections. This could mislead a student into thinking the interval was obtained directly rather than by the required two bisection steps.qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies the Intermediate Value Theorem and performs two steps of bisection to identify the interval [1, 5/4] containing the root.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution claims to perform two steps of bisection but only computes the midpoint of the first interval (3/2) and one endpoint of the second interval (5/4). It fails to explicitly state the second bisection interval [1, 3/2] and its midpoint, making the logical jump to [1, 5/4] incomplete and confusing. Furthermore, f(5/4) is positive, so the root is in [1, 5/4], but the solution does not explicitly verify the sign change on this new interval relative to f(1), relying on the reader to infer it.gpt-oss:20b: pass 2026-10-14
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.