Find a value of c such that the conclusion of the Mean Value Theorem is…
Find a value of c such that the conclusion of the Mean Value Theorem is satisfied for
f(x) = -2x3 + 6x - 2 on the interval [-2, 2]
Answer: A. 2 √(1/3) and -2 √(1/3) — The Mean Value Theorem (MVT) states: if f is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least…
- A.
2 √(1/3) and -2 √(1/3)
- B.
4 √(1/3) and -4 √(1/3)
- C.
√(1/3) and -√(1/3)
- D.
3 √(1/2) and -3 √(1/2)
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Correct answer: A
The Mean Value Theorem (MVT) states: if f is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that f'(c) equals the average rate of change of f over [a, b], i.e. f'(c) = [f(b) - f(a)] / (b - a).
Here f(x) = -2x3 + 6x - 2 is a polynomial, so it is continuous on [-2, 2] and differentiable on (-2, 2) — the MVT applies. Apply it step by step:
Evaluate the endpoints: f(-2) = -2(-2)3 + 6(-2) - 2 = 16 - 12 - 2 = 2
f(2) = -2(2)3 + 6(2) - 2 = -16 + 12 - 2 = -6
Average rate of change: [f(2) - f(-2)] / (2 - (-2)) = (-6 - 2) / 4 = -2
Differentiate: f'(x) = -6x2 + 6
Set f'(c) equal to the average rate of change: -6c2 + 6 = -2
Solve: -6c2 = -8, so c2 = 4/3, giving c = ±√(4/3) = ±2√(1/3)
Both ±2√(1/3) ≈ ±1.15 lie strictly inside the open interval (-2, 2), satisfying the MVT's requirement that c be an interior point. Substituting back: -6(4/3) + 6 = -8 + 6 = -2, which matches the average rate of change computed above.
Hence c = 2√(1/3) and c = -2√(1/3).