Intermediate Value Theorem: locating a root
Problem 1.609 · medium
Show that \( \displaystyle x + e^{x} - 4 = 0 \) 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.
- \[ \left. x + e^{x} - 4 \right|_{\substack{ x=1 }} = -3 + e \]f(1).✓ Proved
- \[ \left. x + e^{x} - 4 \right|_{\substack{ x=2 }} = -2 + e^{2} \]f(2): the opposite sign, so a root lies between.✓ Proved
- \[ \left. x + e^{x} - 4 \right|_{\substack{ x=\frac{3}{2} }} = - \frac{5}{2} + e^{\frac{3}{2}} \]f(3/2) ≈ 1.9817.✓ Proved
- \[ \left. x + e^{x} - 4 \right|_{\substack{ x=\frac{5}{4} }} = - \frac{11}{4} + e^{\frac{5}{4}} \]f(5/4) ≈ 0.7403.✓ 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: passqwen3.6:27b-mlx: fail (error) — The solution claims the sign changes on [1, 5/4], but f(1) is negative and f(5/4) is positive. While a root does lie in this interval, the bisection logic is flawed: since f(1) < 0 and f(1.5) > 0, the root is in [1, 1.5]. The next step should bisect [1, 1.5] to get [1, 1.25] or [1.25, 1.5]. Since f(1.25) > 0, the root is in [1, 1.25]. The conclusion is correct, but the reasoning in step 6 implies checking the sign change between 1 and 1.25 directly without establishing the intermediate interval [1, 1.5] first, skipping the logical step of identifying which sub-interval of the first bisection contains the root. More critically, step 6 says 'The sign changes on [1, 5/4]', which is true, but it fails to explicitly state that the root is in [1, 5/4] *because* f(1) and f(5/4) have opposite signs, which is the correct application of IVT for the second step. However, the real error is that the problem asks for *two* steps of bisection. Step 1: Bisect [1,2] -> [1, 1.5] (since f(1)<0, f(1.5)>0). Step 2: Bisect [1, 1.5] -> [1, 1.25] (since f(1)<0, f(1.25)>0). The solution calculates f(1.5) and f(1.25) but doesn't explicitly structure the two steps. It jumps to the final interval. The sentence 'The sign changes on [1, 5/4]' is technically correct but misleading as a summary of 'two steps of bisection' because it ignores the intermediate interval [1, 1.5] which is the result of the first step. A student might think you just check endpoints and the final midpoint. The correct procedure is: 1. Check mid 1.5. Root in [1, 1.5]. 2. Check mid 1.25. Root in [1, 1.25]. The solution presents the calculations but the narrative sentence 6 is weak/incomplete regarding the process.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution claims the sign changes on [1, 5/4], but f(1) is negative and f(5/4) is positive. While a root does lie in this interval, the bisection logic is flawed: since f(1) < 0 and f(1.5) > 0, the root is in [1, 1.5]. The next step should bisect [1, 1.5] to get [1, 1.25] or [1.25, 1.5]. Since f(1.25) > 0, the root is in [1, 1.25]. The conclusion is correct, but the reasoning in step 6 implies checking the sign change between 1 and 1.25 directly without establishing the intermediate interval [1, 1.5] first, skipping the logical step of identifying which sub-interval of the first bisection contains the root. More critically, step 6 says 'The sign changes on [1, 5/4]', which is true, but it fails to explicitly state that the root is in [1, 5/4] *because* f(1) and f(5/4) have opposite signs, which is the correct application of IVT for the second step. However, the real error is that the problem asks for *two* steps of bisection. Step 1: Bisect [1,2] -> [1, 1.5] (since f(1)<0, f(1.5)>0). Step 2: Bisect [1, 1.5] -> [1, 1.25] (since f(1)<0, f(1.25)>0). The solution calculates f(1.5) and f(1.25) but doesn't explicitly structure the two steps. It jumps to the final interval. The sentence 'The sign changes on [1, 5/4]' is technically correct but misleading as a summary of 'two steps of bisection' because it ignores the intermediate interval [1, 1.5] which is the result of the first step. A student might think you just check endpoints and the final midpoint. The correct procedure is: 1. Check mid 1.5. Root in [1, 1.5]. 2. Check mid 1.25. Root in [1, 1.25]. The solution presents the calculations but the narrative sentence 6 is weak/incomplete regarding the process.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution claims that f(2) has the opposite sign to f(1), but f(1) = e - 3 < 0 and f(2) = e^2 - 2 > 0, so they do have opposite signs, but the text says 'the opposite sign' implying a direct comparison without stating f(1) is negative. More critically, step 6 concludes the root is in [1, 5/4] based on a sign change, but f(1) < 0 and f(5/4) > 0, so the root is indeed in [1, 5/4]. However, the problem asks for two steps of bisection. Step 1 bisects [1,2] to [1.5, 2] or [1, 1.5]. f(1.5) > 0, so root is in [1, 1.5]. Step 2 bisects [1, 1.5] to [1.25, 1.5] or [1, 1.25]. f(1.25) > 0, so root is in [1, 1.25]. The solution correctly identifies the interval [1, 5/4]. Wait, let me re-read carefully. f(1) < 0. f(1.5) > 0. Root in [1, 1.5]. f(1.25) > 0. Root in [1, 1.25]. The interval is [1, 5/4]. The logic holds. Why did I think it failed? Let's check the signs again. f(1) = e - 3 approx -0.28. f(2) = e^2 - 2 approx 5.39. Opposite signs. Correct. f(1.5) = e^1.5 - 2.5 approx 4.48 - 2.5 = 1.98 > 0. So root in [1, 1.5]. f(1.25) = e^1.25 - 2.75 approx 3.49 - 2.75 = 0.74 > 0. So root in [1, 1.25]. The interval is [1, 5/4]. The solution is correct. Is there a subtle error? 'The sign changes on [1, 5/4]'. f(1) is negative, f(5/4) is positive. Yes. The solution is sound.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer returned a non-objectgpt-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.