Home›Calculus 1›The Mean Value Theorem and Rolle's theorem›Problem 3.552

The Mean Value Theorem and Rolle's theorem

Problem 3.552 · hard

Verify that \( \displaystyle f(x) = - 2 x^{3} + 4 x^{2} + 6 x \) satisfies the hypotheses of Rolle's theorem on \( \displaystyle [-1, 3] \), and find every number \( \displaystyle c \) the theorem guarantees.

f is a polynomial, so it is continuous on [a, b] and differentiable on (a, b): the hypotheses hold.
\[ - \frac{\left. - 2 x^{3} + 4 x^{2} + 6 x \right|_{\substack{ x=-1 }}}{4} + \frac{\left. - 2 x^{3} + 4 x^{2} + 6 x \right|_{\substack{ x=3 }}}{4} = 0 \]
The slope of the secant line.✓ Proved
\[ \frac{d}{d x} \left(- 2 x^{3} + 4 x^{2} + 6 x\right) = - 6 x^{2} + 8 x + 6 \]
Differentiate.✓ Proved
\[ - \frac{8 \sqrt{13}}{3} - 6 \left(\frac{2}{3} - \frac{\sqrt{13}}{3}\right)^{2} + \frac{34}{3} = 0 \]
c = 2/3 - sqrt(13)/3 solves f′(c) = 0 and lies in (-1, 3).✓ Proved
\[ - 6 \left(\frac{2}{3} + \frac{\sqrt{13}}{3}\right)^{2} + \frac{8 \sqrt{13}}{3} + \frac{34}{3} = 0 \]
c = 2/3 + sqrt(13)/3 solves f′(c) = 0 and lies in (-1, 3).✓ Proved

Answer \( c = \frac{2}{3} - \frac{\sqrt{13}}{3},\ \frac{2}{3} + \frac{\sqrt{13}}{3} \)

Lines: 4 proved, 1 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.

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
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	each c checked by a difference quotient; a scan of (a, b) finds no others

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to verify the condition f(-1) = f(3), which is required for Rolle's Theorem. Instead, it computes the slope of the secant line (which is 0, implying f(-1)=f(3), but this is not explicitly stated as the check). More critically, it does not explicitly state that f(-1) = f(3) is satisfied, jumping straight to the derivative. While the algebra implies it, the logical step of verifying the endpoint equality is missing from the sentences. Additionally, the problem asks to 'find every number c', and while the roots are found, the solution does not explicitly check if both roots lie within the interval (-1, 3) in the text, though the equations claim they do. The primary error is the lack of explicit verification of f(a)=f(b) in the reasoning steps.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to verify the condition f(-1) = f(3), which is required for Rolle's Theorem. Instead, it computes the slope of the secant line (which is 0, implying f(-1)=f(3), but this is not explicitly stated as the check). More critically, it does not explicitly state that f(-1) = f(3) is satisfied, jumping straight to the derivative. While the algebra implies it, the logical step of verifying the endpoint equality is missing from the sentences. Additionally, the problem asks to 'find every number c', and while the roots are found, the solution does not explicitly check if both roots lie within the interval (-1, 3) in the text, though the equations claim they do. The primary error is the lack of explicit verification of f(a)=f(b) in the reasoning steps.
gpt-oss:20b: pass 2026-10-10
qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to verify the condition f(-1) = f(3), which is required for Rolle's Theorem. Instead, it calculates the slope of the secant line (which is 0, implying f(-1)=f(3)), but does not explicitly state or verify the equality of the function values. More critically, the problem asks to verify the hypotheses, and simply stating 'hypotheses hold' without checking f(a)=f(b) is incomplete. However, the bigger issue is that the solution presents the verification of the roots of f'(x)=0 as if they are the only step, but it misses explicitly confirming f(-1)=f(3) in the text, relying on an unchecked equation for the slope. Actually, looking closer, step 2 calculates the slope as 0. This implies f(3)-f(-1) = 0, so f(3)=f(-1). This is technically correct but presented as a slope calculation rather than a direct check of the hypothesis. The main error is likely that the solution does not explicitly state f(-1) = f(3) as a verified hypothesis, but rather computes the slope. But wait, is there a calculation error? f(-1) = -2(-1)^3 + 4(-1)^2 + 6(-1) = 2 + 4 - 6 = 0. f(3) = -2(27) + 4(9) + 6(3) = -54 + 36 + 18 = 0. So f(-1)=f(3)=0. The slope is 0. The derivative is -6x^2 + 8x + 6. Setting to 0: -3x^2 + 4x + 3 = 0 => 3x^2 - 4x - 3 = 0. x = (4 +/- sqrt(16 + 36))/6 = (4 +/- sqrt(52))/6 = (4 +/- 2sqrt(13))/6 = 2/3 +/- sqrt(13)/3. The roots are correct. The interval is (-1, 3). sqrt(13) is approx 3.6. 2/3 - 1.2 = -0.53 (in interval). 2/3 + 1.2 = 1.86 (in interval). The math is correct. Why would it fail? The prompt says 'Verify that ... satisfies the hypotheses'. Step 1 says 'hypotheses hold'. Step 2 checks the slope is 0. This is sufficient to verify f(a)=f(b). Is there a subtle error? The problem asks to find 'every number c'. The solution finds two. Both are in (-1, 3). The solution seems correct. Let me re-read the instructions. 'Check that ... each conclusion follows from the lines before it'. Step 1 claims hypotheses hold. Step 2 shows slope is 0. This supports the third hypothesis. Continuity and differentiability are standard for polynomials. The solution is sound. Wait, is the interval open or closed for differentiability? Differentiable on (a,b). Continuous on [a,b]. Step 1 says continuous on [a,b] and differentiable on (a,b). This is correct. Why would it be a fail? Maybe the 'unchecked' equations are not proofs? The prompt says 'do not re-check the algebra of a line marked proved'. These are marked 'unchecked'. But the instructions say 'Each line ... is either an EQUATION ... or a SENTENCE'. The review is for the SENTENCES. Step 1 is a sentence. It claims hypotheses hold. Does it follow? It relies on Step 2 implicitly? No, Step 1 is a standalone claim. A rigorous solution should show f(a)=f(b). Step 2 does this via slope. But Step 1 makes the claim before Step 2 provides the evidence for the specific condition f(a)=f(b). This is a logical ordering issue or a lack of explicit verification in the text. However, usually 'pass' is for sound solutions. Is it misleading? No. Is it an error? The claim 'hypotheses hold' is made without explicitly verifying f(a)=f(b) in the text of step 1, but step 2 does it. It's a bit disjointed. Let's look for a harder error. Did it check that c is in (-1, 3)? Steps 4 and 5 say 'lies in (-1, 3)'. This is a sentence part of the equation comment. It asserts it. Is it true? Yes. Is it verified? No, just asserted. A student might not know how to check. But the prompt asks to review the solution. If the solution asserts something true without proof, is it an error? In calculus, checking if a root is in an interval is often done by estimation. The solution doesn't show the estimation. It just states it. This might be considered 'style' or 'misleading' if it implies it's obvious. But let's look at the verdict options. 'error' is for wrong steps. 'misleading' for false learning. 'style' for clarity. The solution is mathematically correct. The only potential issue is the lack of explicit verification of the interval containment for c. However, the roots are clearly within the bounds. I will mark it as pass because the math is correct and the steps are logically sufficient, even if brief.
gpt-oss:20b: pass 2026-10-10

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/mean_value_theorem, checked 2026-10-10 with SymPy 1.14.0.