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[` (1) `2f^'(c)""=g^'(c)` (2) `2f^'(c)""=""3g^'(c)` (3) `f^'(c)""=g^'(c)` (4) `f'(c)""=""2g'(c)`

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 the functions \( f \) and \( g \) on the interval \([0, 1]\).

### Step 1: Apply the Mean Value Theorem to \( f(x) \)

According to the Mean Value Theorem, if \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that:

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

For \( f \), we have:
- \( a = 0 \)
- \( b = 1 \)
- \( f(0) = 2 \)
- \( f(1) = 6 \)

So, we can calculate:

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

### Step 2: Apply the Mean Value Theorem to \( g(x) \)

Now, applying the Mean Value Theorem to \( g \):

For \( g \), we have:
- \( g(0) = 0 \)
- \( g(1) = 2 \)

So we can calculate:

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

### Step 3: Relate \( f'(c) \) and \( g'(c) \)

From the results of the Mean Value Theorem, we have:

1. \( f'(c) = 4 \)
2. \( g'(c) = 2 \)

Now, we can relate \( f'(c) \) and \( g'(c) \):

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

This means that:

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

### Conclusion

Thus, we have shown that:

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

Based on the options provided, the correct answer is:

**(4) \( f'(c) = 2g'(c) \)**