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

Given that f(2) = 2 and f '(x) ≥ 1 for 2 ≤ x ≤ 7, we can use the Mean Value Theorem to find the smallest possible value for f(7).

The Mean Value Theorem states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists some c in the interval (a, b) such that f '(c) = (f(b) - f(a)) / (b - a).

In this case, we know that f '(x) ≥ 1 for 2 ≤ x ≤ 7. This means that the average rate of change of the function from x = 2 to x = 7 is at least 1.

So, we can write the inequality (f(7) - f(2)) / (7 - 2) ≥ 1.

Substituting f(2) = 2, we get (f(7) - 2) / 5 ≥ 1.

Solving for f(7), we find that f(7) ≥ 7.

Therefore, the smallest possible value for f(7) is 7.