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F f ( 2 ) = 14 and f ' ( x ) ≥ 1 for 2 ≤ x ≤ 7 , how small can f ( 7 ) possibly be?

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Answer:

f(7) ≥ 19

Step-by-step explanation:To find the smallest possible value of f(7), we can use the Mean Value Theorem for Derivatives. According to this theorem, if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that:

f'(c) = (f(b) - f(a))/(b - a)

In this case, we know that f(2) = 14 and f'(x) ≥ 1 for 2 ≤ x ≤ 7. Therefore, we can apply the Mean Value Theorem to the interval [2, 7] to get:

f'(c) = (f(7) - f(2))/(7 - 2)

Since f'(x) ≥ 1 for 2 ≤ x ≤ 7, we have:

1 ≤ f'(c) = (f(7) - 14)/5

Multiplying both sides by 5 and adding 14, we get:

f(7) ≥ 19

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