Understanding the Problem: Differentiable Functions and Ratios

The question asks for the value of the ratio of derivatives $ \frac{f'(c)}{g'(c)} $ at a specific point $ c $ within the interval $ [0, 1] $. We are given that $ f(x) $ and $ g(x) $ are differentiable functions on $ 0 \le x \le 1 $. We also have a specific relationship between the function values at the endpoints: $ f(1)-f(0) = k(g(1)-g(0)) $, with $ k \ne 0 $. The existence of a point $ c $ such that $ 0 < c < 1 $ is guaranteed.

This setup strongly suggests the use of theorems related to the Mean Value Theorem, particularly Cauchy's Mean Value Theorem, or a variation involving Rolle's Theorem.

Applying Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem states that if two functions, say $ \phi(x) $ and $ \psi(x) $, are continuous on the closed interval $ [a, b] $ and differentiable on the open interval $ (a, b) $, and $ \psi'(x) \ne 0 $ for all $ x \in (a, b) $, then there exists at least one number $ c \in (a, b) $ such that:

In our problem:

• The interval is $ [a, b] = [0, 1] $.
• The functions are $ f(x) $ and $ g(x) $.
• We are given that $ f(x) $ and $ g(x) $ are differentiable on $ [0, 1] $. This implies they are also continuous on $ [0, 1] $ and differentiable on $ (0, 1) $.
• We are given the condition $ f(1)-f(0) = k(g(1)-g(0)) $.

Let's apply Cauchy's MVT with $ \phi(x) = f(x) $ and $ \psi(x) = g(x) $. According to the theorem, there exists a $ c \in (0, 1) $ such that:

Now, we use the given condition $ f(1)-f(0) = k(g(1)-g(0)) $. Substitute this into the equation derived from Cauchy's MVT:

Assuming $ g(1)-g(0) \ne 0 $, we can cancel the term $ (g(1)-g(0)) $. The problem implies the existence of such a $ c $ and the ratio is sought, which usually means the conditions for the theorem apply and the denominator $ g'(c) $ is non-zero.

Alternative Explanation using Rolle's Theorem

We can also solve this by constructing an auxiliary function based on the given condition and applying Rolle's Theorem. Rolle's Theorem states that if a function $ H(x) $ is continuous on $ [a, b] $, differentiable on $ (a, b) $, and $ H(a) = H(b) $, then there exists at least one number $ c \in (a, b) $ such that $ H'(c) = 0 $.

Let's define a new function $ H(x) = f(x) - \lambda g(x) $ for some constant $ \lambda $. We need to choose $ \lambda $ such that $ H(0) = H(1) $.

• $ H(1) = f(1) - \lambda g(1) $
• $ H(0) = f(0) - \lambda g(0) $

Setting $ H(1) = H(0) $:

Rearranging the terms:

We are given the condition $ f(1) - f(0) = k(g(1) - g(0)) $. Comparing these two equations, we find that we must choose $ \lambda = k $.

Therefore, let's consider the function $ H(x) = f(x) - k g(x) $. Since $ f(x) $ and $ g(x) $ are differentiable on $ [0, 1] $, $ H(x) $ is also differentiable on $ [0, 1] $ and continuous on $ [0, 1] $. We have shown that $ H(0) = H(1) $.

By Rolle's Theorem, there exists a $ c \in (0, 1) $ such that $ H'(c) = 0 $.

Let's find $ H'(x) $:

Now, set $ H'(c) = 0 $:

Rearrange the equation:

To find the ratio $ \frac{f'(c)}{g'(c)} $, we divide both sides by $ g'(c) $, assuming $ g'(c) \ne 0 $:

Conclusion

Both methods, applying Cauchy's Mean Value Theorem directly and using Rolle's Theorem with an auxiliary function, lead to the same result. The value of $ \frac{f'(c)}{g'(c)} $ is $ k $.