The f represents Lagrange’s Mean Value Theorem. Before, the representation of the theorem, it is important to understand what does the mean value theorem represents. The mean value theorem of Lagrange represents that if a function f (x), is continuous, within a closed interval, that is, [a, b], and it is differentiable, once placed on an open interval, that is, (a, b), and there will always be one certain point, which will be placed in the interval, that is c= x. therefore, the theorem represents,
f (b) – f (a) = f” (c) (b – a).
This particular theorem therefore, helps in the expressing the increment of a function in an interval, through the expression of the value of the derivative at the intermediate point.
Now, according to the problem, the f (x), satisfies the requirements of the Mean Value Theorem of Lagrange, therefore,
f is Lagrange’s Mean Value Theorem which is a in a closed interval of [0, 2].
Now, f (0) = 0 and,
f’ (x) ≤ ½.
Now, £ c € [0, 2], is subjected to,
f’ (c) = f (2) – f (0)/ 2 – 0
Therefore, f’ (c) = f (2)/ 2
Therefore, since, f (x) ≤ ½, therefore,
f’ (c) ≤ ½.