Intermediate Value Theorem: locating a root
Problem 1.605 · medium
Show that \( \displaystyle x^{3} - 7 = 0 \) 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.
- \[ \left. x^{3} - 7 \right|_{\substack{ x=1 }} = -6 \]f(1).✓ Proved
- \[ \left. x^{3} - 7 \right|_{\substack{ x=2 }} = 1 \]f(2): the opposite sign, so a root lies between.✓ Proved
- \[ \left. x^{3} - 7 \right|_{\substack{ x=\frac{3}{2} }} = - \frac{29}{8} \]f(3/2) ≈ -3.6250.✓ Proved
- \[ \left. x^{3} - 7 \right|_{\substack{ x=\frac{7}{4} }} = - \frac{105}{64} \]f(7/4) ≈ -1.6406.✓ Proved
- The sign changes on [7/4, 2], so a root lies there.
Answer \( \left[\frac{7}{4}, 2\right] \)
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 first bisection step (at x=3/2) and jumps to 7/4 without explaining why 7/4 is chosen. It therefore does not correctly demonstrate two bisection steps, making the reasoning misleading.qwen3.6:27b-mlx: fail (error) — The solution performs only one step of bisection (checking the midpoint 3/2) but claims to use two steps. A second step is required to bisect the interval [3/2, 2] and check the midpoint 7/4, which is done, but the logic skips the explicit identification of the first sub-interval [3/2, 2] containing the root before proceeding to the second bisection. More critically, the final interval [7/4, 2] has length 1/4, which is correct, but the narrative implies the root is in [7/4, 2] based on the sign change between 7/4 and 2. However, f(7/4) is negative and f(2) is positive, so the root is indeed in [7/4, 2]. Wait, let's re-read carefully. Step 4 checks 3/2. f(3/2) < 0, f(2) > 0. So root is in [3/2, 2]. Step 5 checks 7/4 (midpoint of 3/2 and 2). f(7/4) < 0, f(2) > 0. So root is in [7/4, 2]. The solution lists the calculations but the sentence in step 6 says 'The sign changes on [7/4, 2]'. This is true. Did it use two steps? Yes: 1. Bisect [1,2] -> [3/2, 2]. 2. Bisect [3/2, 2] -> [7/4, 2]. The solution is actually correct in its result and steps. Why did I think it failed? Let's look closer. Step 4 calculates f(3/2). Step 5 calculates f(7/4). Step 6 concludes root is in [7/4, 2]. This is logically sound. Is there a missing sentence? The prompt asks to 'use two steps of bisection'. The solution shows the evaluations. It doesn't explicitly state 'Since f(3/2)<0 and f(2)>0, the root is in [3/2, 2]'. It jumps to checking 7/4. While the math is right, the explanation is slightly compressed. However, is it an error? No. Is it misleading? Maybe slightly, but a student would likely follow. Let's look for a harder error. f(1)=-6, f(2)=1. Midpoint 1.5. f(1.5) = 3.375 - 7 = -3.625. Negative. Root in [1.5, 2]. Midpoint of [1.5, 2] is 1.75 = 7/4. f(1.75) = 1.75^3 - 7 = 5.359375 - 7 = -1.640625. Negative. Root in [1.75, 2]. Interval [7/4, 2]. Length 0.25. The solution is correct. Wait, look at step 6: 'The sign changes on [7/4, 2]'. This implies f(7/4) and f(2) have opposite signs. f(7/4) is negative, f(2) is positive. This is correct. The solution is sound. Why would I flag it? Perhaps the lack of explicit intermediate conclusion? 'Step 1: Root in [3/2, 2]. Step 2: Root in [7/4, 2].' The solution just lists equations. Sentence 6 summarizes the final state. It is a bit terse but not wrong. Let's check if there's a subtle error. 'Subs(x**3 - 7, x, 7/4) = -105/64'. 7/4 = 1.75. 1.75^3 = (7/4)^3 = 343/64. 343/64 - 7 = 343/64 - 448/64 = -105/64. Correct. The verdict should be pass. Wait, I need to be adversarial. Is there anything wrong? The problem asks to 'Show that ... has a solution ... then use two steps'. The solution shows IVT applies. Then it does the steps. The final answer is [7/4, 2]. This is correct. I will mark it as pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution performs only one step of bisection (checking the midpoint 3/2) but claims to use two steps. A second step is required to bisect the interval [3/2, 2] and check the midpoint 7/4, which is done, but the logic skips the explicit identification of the first sub-interval [3/2, 2] containing the root before proceeding to the second bisection. More critically, the final interval [7/4, 2] has length 1/4, which is correct, but the narrative implies the root is in [7/4, 2] based on the sign change between 7/4 and 2. However, f(7/4) is negative and f(2) is positive, so the root is indeed in [7/4, 2]. Wait, let's re-read carefully. Step 4 checks 3/2. f(3/2) < 0, f(2) > 0. So root is in [3/2, 2]. Step 5 checks 7/4 (midpoint of 3/2 and 2). f(7/4) < 0, f(2) > 0. So root is in [7/4, 2]. The solution lists the calculations but the sentence in step 6 says 'The sign changes on [7/4, 2]'. This is true. Did it use two steps? Yes: 1. Bisect [1,2] -> [3/2, 2]. 2. Bisect [3/2, 2] -> [7/4, 2]. The solution is actually correct in its result and steps. Why did I think it failed? Let's look closer. Step 4 calculates f(3/2). Step 5 calculates f(7/4). Step 6 concludes root is in [7/4, 2]. This is logically sound. Is there a missing sentence? The prompt asks to 'use two steps of bisection'. The solution shows the evaluations. It doesn't explicitly state 'Since f(3/2)<0 and f(2)>0, the root is in [3/2, 2]'. It jumps to checking 7/4. While the math is right, the explanation is slightly compressed. However, is it an error? No. Is it misleading? Maybe slightly, but a student would likely follow. Let's look for a harder error. f(1)=-6, f(2)=1. Midpoint 1.5. f(1.5) = 3.375 - 7 = -3.625. Negative. Root in [1.5, 2]. Midpoint of [1.5, 2] is 1.75 = 7/4. f(1.75) = 1.75^3 - 7 = 5.359375 - 7 = -1.640625. Negative. Root in [1.75, 2]. Interval [7/4, 2]. Length 0.25. The solution is correct. Wait, look at step 6: 'The sign changes on [7/4, 2]'. This implies f(7/4) and f(2) have opposite signs. f(7/4) is negative, f(2) is positive. This is correct. The solution is sound. Why would I flag it? Perhaps the lack of explicit intermediate conclusion? 'Step 1: Root in [3/2, 2]. Step 2: Root in [7/4, 2].' The solution just lists equations. Sentence 6 summarizes the final state. It is a bit terse but not wrong. Let's check if there's a subtle error. 'Subs(x**3 - 7, x, 7/4) = -105/64'. 7/4 = 1.75. 1.75^3 = (7/4)^3 = 343/64. 343/64 - 7 = 343/64 - 448/64 = -105/64. Correct. The verdict should be pass. Wait, I need to be adversarial. Is there anything wrong? The problem asks to 'Show that ... has a solution ... then use two steps'. The solution shows IVT applies. Then it does the steps. The final answer is [7/4, 2]. This is correct. I will mark it as pass.gpt-oss:20b: fail (misleading) 2026-10-17 — The solution skips the first bisection step (at x=3/2) and jumps to 7/4 without explaining why 7/4 is chosen. It therefore does not correctly demonstrate two bisection steps, making the reasoning misleading.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to explicitly state that f(2) is positive, which is required to establish the sign change on [7/4, 2] via the Intermediate Value Theorem. While f(2)=1 was calculated in step 3, step 6 does not reference this fact, making the logical deduction incomplete.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-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 skips the required two bisection steps and merely asserts that the root lies in [7/4,2] without demonstrating the intermediate sign changes that justify each bisection. This omission means the method is not correctly applied and the student would learn an incomplete 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.