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Intermediate Value Theorem: locating a root

Problem 1.611 · medium

Show that \( \displaystyle x^{4} + x - 3 = 0 \) has a solution in \( \displaystyle [-2, -1] \), then use two steps of bisection to find an interval of length \( \displaystyle \tfrac14 \) containing it.

f is continuous on [-2, -1] (it is built from continuous functions), so the IVT applies.
\[ \left. x^{4} + x - 3 \right|_{\substack{ x=-2 }} = 11 \]
f(-2).✓ Proved
\[ \left. x^{4} + x - 3 \right|_{\substack{ x=-1 }} = -3 \]
f(-1): the opposite sign, so a root lies between.✓ Proved
\[ \left. x^{4} + x - 3 \right|_{\substack{ x=- \frac{3}{2} }} = \frac{9}{16} \]
f(-3/2) ≈ 0.5625.✓ Proved
\[ \left. x^{4} + x - 3 \right|_{\substack{ x=- \frac{5}{4} }} = - \frac{463}{256} \]
f(-5/4) ≈ -1.8086.✓ Proved
The sign changes on [-3/2, -5/4], so a root lies there.

Answer \( \left[- \frac{3}{2}, - \frac{5}{4}\right] \)

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: pass
qwen3.6:27b-mlx: fail (error) — The interval notation in step 6 is incorrect because the left endpoint (-1.5) is greater than the right endpoint (-1.25). Standard interval notation requires [a, b] with a <= b, so it should be [-5/4, -3/2].

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-17 — The interval notation in step 6 is incorrect because the left endpoint (-1.5) is greater than the right endpoint (-1.25). Standard interval notation requires [a, b] with a <= b, so it should be [-5/4, -3/2].
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The interval notation in step 6 is invalid because the lower bound (-3/2) is greater than the upper bound (-5/4). Standard interval notation requires [a, b] with a <= b. The correct interval is [-5/4, -3/2] or [-1.25, -1.5] is wrong, it should be [-1.5, -1.25]. Wait, -3/2 = -1.5 and -5/4 = -1.25. So -1.5 < -1.25. The interval [-1.5, -1.25] is valid. Let me re-read step 6. It says [-3/2, -5/4]. -3/2 is -1.5. -5/4 is -1.25. So the interval is [-1.5, -1.25]. This is a valid interval. Why did I think it was invalid? Ah, I misread the order in my head. -1.5 is less than -1.25. So [-1.5, -1.25] is correct. Let me re-evaluate the bisection steps. Initial interval: [-2, -1]. Length 1. Midpoint: -1.5 (-3/2). f(-2) = 11 (>0). f(-1) = -3 (<0). f(-1.5) = 9/16 (>0). Since f(-1.5) > 0 and f(-1) < 0, the root is in [-1.5, -1]. This is the first bisection step. Second bisection step: Interval: [-1.5, -1]. Midpoint: (-1.5 + -1)/2 = -2.5/2 = -1.25 (-5/4). f(-1.5) = 9/16 (>0). f(-1.25) = -463/256 (<0). Since f(-1.5) > 0 and f(-1.25) < 0, the root is in [-1.5, -1.25]. This interval is [-3/2, -5/4]. Length: |-1.25 - (-1.5)| = 0.25 = 1/4. The solution is actually correct. The interval [-3/2, -5/4] is [-1.5, -1.25], which is a valid interval where the lower bound is less than the upper bound. My initial quick check was flawed. The steps are correct. The signs are correct. The interval selection is correct. The final answer matches the stated answer. Wait, let me double check the problem statement. "find an interval of length 1/4 containing it". The interval [-3/2, -5/4] has length 1/4. It contains the root. The steps show the bisection process correctly. Step 6 concludes the root lies in that interval. This is correct. So why did I flag it? I didn't flag it yet. I am generating the response. The solution is sound. Let me look closer at Step 6. "The sign changes on [-3/2, -5/4]". f(-3/2) is positive. f(-5/4) is negative. Yes, sign changes. So a root lies there. This is correct. Is there any subtle error? Step 1: IVT applies. Correct. Step 2: f(-2)=11. Correct. Step 3: f(-1)=-3. Correct. Root in [-2, -1]. Step 4: f(-3/2) = 9/16. Correct. Step 5: f(-5/4) = -463/256. Correct. Step 6: Root in [-3/2, -5/4]. Correct. The solution is perfect.
gpt-oss:20b: fail (misleading) 2026-10-14 — The solution skips the required bisection steps: after evaluating at -3/2 it should test the midpoint of [-3/2,-1] (i.e., -11/8), not jump to -5/4. Thus the conclusion that the root lies in [-3/2,-5/4] is not justified by two bisection steps.
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The interval notation [-3/2, -5/4] is invalid because the lower bound (-1.5) is greater than the upper bound (-1.25). The correct interval is [-5/4, -3/2].
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.