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Consider a function f: [−7, 0] → R that is continuous on [−7, 0] and differentiable on (−7, 0). Given that f(−7) = −3 and f'(x) ≤ 2 for every x in (−7, 0), the quantity f'(−1) + f(0) must belong to which interval for all such functions?
- (−∞, 20]
- [−3, 11]
- (−∞, 11]
- [−6, 20]
Correct answer: (−∞, 20]
Solution
By MVT, f(0) <= f(-7)+2*7 = 11, and f'(-1) <= 2, so f'(-1)+f(0) can reach 13 (take f'=2 everywhere, f(x)=2x+11). Since f' has no lower bound, the sum is unbounded below, so it must lie in (-inf, 20].
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