Question

If $f$ and $g$ are differentiable functions in $(0, 1)$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$, then for some $c \in ] 0, 1 [$

• A. $2f ' (c) = g ' (c)$
• B. $2f ' (c) = 3 g ' (c)$
• C. $f ' (c) = g ' (c)$
• D. $f ' (c) = 2 g ' (c)$

Answer 1

The Correct Option is D

To solve this problem, we will use the Mean Value Theorem (MVT) for derivatives which states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c \in (a, b) such that:

f'(c) = \frac{f(b) - f(a)}{b - a}

Similarly, the conditions on the functions f and g allow us to apply MVT to them.

1. For the function f on interval [0, 1]:

   We have f(0) = 2 and f(1) = 6. Using MVT, there exists c_1 \in (0, 1) such that:

   f'(c_1) = \frac{f(1) - f(0)}{1 - 0} = \frac{6 - 2}{1 - 0} = 4

2. For the function g on interval [0, 1]:

   We have g(0) = 0 and g(1) = 2. Using MVT, there exists c_2 \in (0, 1) such that:

   g'(c_2) = \frac{g(1) - g(0)}{1 - 0} = \frac{2 - 0}{1 - 0} = 2

Given the options, f'(c) = 2g'(c) is the correct answer because it reflects the relationship we derived using the Mean Value Theorem:

• From the MVT on f, f'(c_1) = 4.
• From the MVT on g, g'(c_2) = 2.
• Therefore, 4 = 2 \times 2 suggests f'(c) = 2g'(c).

The correct answer thus is: f'(c) = 2g'(c).