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

Options

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

Answer

To solve the problem, we will apply the Mean Value Theorem (MVT) for functions. The MVT states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that:

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

Given the functions \( f \) and \( g \) with the following values:
- \( f(0) = 2 \)
- \( f(1) = 6 \)
- \( g(0) = 0 \)
- \( g(1) = 2 \)

We will apply the MVT to both functions \( f \) and \( g \).

### Step 1: Apply MVT to function \( f \)
Using the MVT for \( f \) on the interval [0, 1]:

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

So, there exists a point \( c_1 \in (0, 1) \) such that:

\[ f'(c_1) = 4 \]

### Step 2: Apply MVT to function \( g \)
Now, applying the MVT for \( g \) on the interval [0, 1]:

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

So, there exists a point \( c_2 \in (0, 1) \) such that:

\[ g'(c_2) = 2 \]

### Step 3: Relate \( f' \) and \( g' \)
From the previous steps, we have:

\[ f'(c_1) = 4 \]
\[ g'(c_2) = 2 \]

We can now relate the derivatives at some point \( c \) in (0, 1). Since we are looking for a relationship between \( f' \) and \( g' \), we can express \( f' \) in terms of \( g' \):

From the information given, we can assume a linear relationship:

\[ f'(c) = 2 \cdot g'(c) \]

This means:

\[ 4 = 2 \cdot g'(c) \]

### Step 4: Solve for \( g'(c) \)
From the equation:

\[ 4 = 2 \cdot g'(c) \]

We can solve for \( g'(c) \):

\[ g'(c) = \frac{4}{2} = 2 \]

### Conclusion
Thus, we have shown that there exists some \( c \in (0, 1) \) such that:

\[ f'(c) = 4 \]
\[ g'(c) = 2 \]

This satisfies the conditions of the Mean Value Theorem for both functions.