Suppose that \(f: \mathbb{R} \rightarrow \mathbb{R}\) is a continuous function…
2021
Suppose that \(f: \mathbb{R} \rightarrow \mathbb{R}\) is a continuous function on the interval [−3,3] and a differentiable function in the interval (−3,3) such that for every \(x\) in the interval, \(f’(x) \leq 2\) if \(f(-3)=7\), then \(f(3)\) is at most __________ .
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Correct answer: 19
Key idea: apply the Mean Value Theorem to the interval [−3,3].
By the Mean Value Theorem there exists some c in (−3,3) with f'(c) = (f(3) − f(−3)) / (3 − (−3)) = (f(3) − 7) / 6.
Given f'(x) ≤ 2 for every x, in particular f'(c) ≤ 2, so (f(3) − 7) / 6 ≤ 2.
Multiply both sides by 6: f(3) − 7 ≤ 12, hence f(3) ≤ 19.
Conclusion: f(3) is at most 19.