Final answer:
Using the Mean Value Theorem, we've shown the required inequality for the function f(x) = sqrt(x) over the interval (x, y) when 0 < x < y, confirming that option c. x < y is correct.
Step-by-step explanation:
The question asks us to use the Mean Value Theorem (MVT) to show that sqrt(y) - sqrt(x) < (y - x)/(2sqrt(x)) for 0 < x < y.
The MVT states that if f(x) is a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists some c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).
To apply the MVT to this problem, we'll consider f(x) = sqrt(x) which is continuous and differentiable on the interval (x, y) with y > x > 0.
Computing the derivative, we get f'(x) = 1/(2sqrt(x)). By the MVT, there exists a c between x and y such that
f'(c) = (f(y) - f(x)) / (y - x) = (sqrt(y) - sqrt(x)) / (y - x).
Hence, we have 1/(2sqrt(c)) = (sqrt(y) - sqrt(x)) / (y - x).
Because c > x, we know that sqrt(c) > sqrt(x), and hence 1/(2sqrt(c)) < 1/(2sqrt(x)).
Multiplying both sides by (y - x), we obtain the inequality sqrt(y) - sqrt(x) < (y - x)/(2sqrt(x)).
Option (c) x < y is the correct option for the final answer because it summarizes the interval condition under which the Mean Value Theorem was applied and the inequality was established.