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Let f: R → R be a continuously differentiable function such that f(2) = 6 and f'(2) = 1/48. If ∫₆^(f(x)) 4t³ dt = (x−2)g(x), then Lim x→2 g(x) is equal to:
- 18
- 36
- 12
- 24
Correct answer: 18
Solution
To find the limit as x approaches 2 of g(x), we can differentiate both sides of the equation using the Fundamental Theorem of Calculus and the Chain Rule. Evaluating at x = 2 gives us g(2) = 4(6³) / f'(2), which simplifies to 18 when substituting the known values of f(2) and f'(2).
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