∫Calc Practice

Intermediate Value Theorem: locating a root

Problem 1.619 · medium

Show that \( \displaystyle x^{5} + x^{2} - 1 = 0 \) has a solution in \( \displaystyle [0, 1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.
  1. f is continuous on [0, 1] (it is built from continuous functions), so the IVT applies.
    Reviewed
  2. \[ \left. x^{5} + x^{2} - 1 \right|_{\substack{ x=0 }} = -1 \]
    f(0).✓ Proved
  3. \[ \left. x^{5} + x^{2} - 1 \right|_{\substack{ x=1 }} = 1 \]
    f(1): the opposite sign, so a root lies between.✓ Proved
  4. \[ \left. x^{5} + x^{2} - 1 \right|_{\substack{ x=1 \cdot \frac{1}{2} }} = - \frac{23}{32} \]
    f(1/2) ≈ -0.7188.✓ Proved
  5. \[ \left. x^{5} + x^{2} - 1 \right|_{\substack{ x=\frac{3}{4} }} = - \frac{205}{1024} \]
    f(3/4) ≈ -0.2002.✓ Proved
  6. The sign changes on [3/4, 1], so a root lies there.
    Reviewed
Answer \( \left[\frac{3}{4}, 1\right] \)

Lines: 4 proved, 2 reviewed. 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
LineStatusChecked byDetail
1Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
6Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0mpmath's bisection finds the root inside the claimed interval

Reviewers

  • gpt-oss:20b: pass
  • 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 interval [3/4, 1].
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 interval [3/4, 1].
  • gpt-oss:20b: pass 2026-10-17
  • 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 an interval of length 1/4.
  • gpt-oss:20b: fail (error) 2026-10-14 — The solution states that a root lies in [3/4,1] but does not perform the two required bisection steps to reduce the interval to length 1/4. The conclusion about the interval is therefore incomplete and 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 interval [3/4, 1], which has the required length of 1/4.
  • 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 2026-10-17 with SymPy 1.14.0.