Suppose that f(0) = 4 and f'(x)< or equal to 2 for x>0. Apply the MVT to the interval [0 , 3] to prove that f(3) < or equal to 10

The mean value theorem states that there is a point x in the interval for which

f'(x) = [f(x2) - f(x1)]/(x2-x1)

With f'(x) < 2 for the entire interval,
[f(x2) - f(x1)]/(x2-x1) cannot exceed 2,
which means
f(x2) - f(x1) < 2*3 = 6
f(x2) = f(3) < 6 + f(x1) = 10
f(3) < 10

To apply the Mean Value Theorem (MVT), we need to verify two conditions:

1. f(x) is continuous on the closed interval [0, 3]
2. f(x) is differentiable on the open interval (0, 3)

According to the given information, f(0) = 4, which satisfies the first condition.

Now, let's examine the second condition. We are given that f'(x) ≤ 2 for x > 0. This implies that f(x) is differentiable on the open interval (0, 3), since the only point of concern is x = 0, but we already know the value of f(0).

Since both conditions of the MVT are satisfied, we can conclude that there exists at least one point c in the open interval (0, 3) such that:

f'(c) = (f(3) - f(0))/(3 - 0)

Let's solve this equation for f(3):

f(3) - f(0) = f'(c) * 3

Since f'(x) ≤ 2 for x > 0, we can set f'(c) ≤ 2:

f(3) - f(0) ≤ 2 * 3
f(3) - 4 ≤ 6
f(3) ≤ 10

Thus, we have proved that f(3) ≤ 10 using the Mean Value Theorem (MVT) on the interval [0, 3].