Key Concepts and Formulas

• Rolle's Theorem: If a function $h(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $h(a) = h(b)$, then there exists a $c \in (a, b)$ such that $h'(c) = 0$.
• Mean Value Theorem (MVT): If a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
• Differentiability implies continuity.

Step-by-Step Solution

Step 1: Construct an Auxiliary Function

We need to construct a function $h(x)$ such that $h(0) = h(1)$ so that we can apply Rolle's Theorem. Let's try a linear combination of $f(x)$ and $g(x)$: $h(x) = f(x) - g(x)$ We want to find a constant $k$ such that $h(x) = f(x) - k g(x)$ and $h(0) = h(1)$.

Step 2: Evaluate h(0) and h(1)

We have: $h(0) = f(0) - k g(0) = 2 - k(0) = 2$ $h(1) = f(1) - k g(1) = 6 - k(2) = 6 - 2k$

Step 3: Determine the Value of k

To apply Rolle's theorem, we need $h(0) = h(1)$. Therefore, $2 = 6 - 2k$ $2k = 4$ $k = 2$

Step 4: Define the Auxiliary Function h(x)

Now we can define our auxiliary function: $h(x) = f(x) - 2g(x)$

Step 5: Verify Rolle's Theorem Conditions

Since $f(x)$ and $g(x)$ are differentiable on $[0, 1]$, $h(x) = f(x) - 2g(x)$ is also differentiable on $[0, 1]$. Differentiability implies continuity, so $h(x)$ is continuous on $[0, 1]$. Also, $h(0) = f(0) - 2g(0) = 2 - 2(0) = 2$ and $h(1) = f(1) - 2g(1) = 6 - 2(2) = 6 - 4 = 2$. Thus, $h(0) = h(1)$. Therefore, all the conditions of Rolle's Theorem are satisfied.

Step 6: Apply Rolle's Theorem

Since $h(x)$ satisfies the conditions of Rolle's Theorem on $[0, 1]$, there exists a $c \in (0, 1)$ such that $h'(c) = 0$.

Step 7: Calculate h'(x) and h'(c)

We have $h(x) = f(x) - 2g(x)$, so $h'(x) = f'(x) - 2g'(x)$ Therefore, $h'(c) = f'(c) - 2g'(c) = 0$

Step 8: Find the Relationship Between f'(c) and g'(c)

From $f'(c) - 2g'(c) = 0$, we get $f'(c) = 2g'(c)$

Common Mistakes & Tips

• A common mistake is to not construct the correct auxiliary function. The key is to find a function $h(x)$ such that $h(a) = h(b)$ to apply Rolle's Theorem.
• Remember that differentiability implies continuity. This is important for verifying the conditions of Rolle's Theorem and the Mean Value Theorem.
• Don't forget to verify that all conditions of Rolle's Theorem are satisfied before applying it.

Summary

We constructed an auxiliary function $h(x) = f(x) - 2g(x)$. We showed that $h(0) = h(1)$, and since $f$ and $g$ are differentiable, $h$ is also differentiable. Therefore, we could apply Rolle's Theorem, which states that there exists a $c \in (0, 1)$ such that $h'(c) = 0$. This gave us the relationship $f'(c) = 2g'(c)$.

Final Answer

The final answer is \boxed{f'(c) = 2g'(c)}, which corresponds to option (B).