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Problem 123

# State and prove the Cauchy Mean Value Theorem.

Expert verified
The Cauchy Mean Value Theorem states that for continuous functions $$f$$ and $$g$$ on $$[a, b]$$ and differentiable on $$(a, b)$$, with $$g'(x) \neq 0$$ for all $$x$$ in $$(a, b)$$, there exists at least one $$c$$ in $$(a, b)$$ such that $$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$. To prove this, we construct a helper function $$h(x) = f(x) - \frac{f(b) - f(a)}{g(b) - g(a)} \cdot g(x)$$, which is continuous and differentiable on $$(a, b)$$. Applying Rolle's theorem to $$h(x)$$ since $$h(a) = h(b)$$, there exists a $$c$$ in $$(a, b)$$ such that $$h'(c) = 0$$. Computing the derivative of the helper function and setting it equal to 0, we obtain $$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$, which proves the Cauchy Mean Value Theorem.
See the step by step solution

## Step 1: State the Cauchy Mean Value Theorem

Let $$f$$ and $$g$$ be continuous functions on the closed interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$. If $$g'(x) \neq 0$$ for all $$x$$ in $$(a, b)$$, then there exists at least one $$c$$ in $$(a, b)$$ such that: $\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$

## Step 2: Introduce a helper function

To help us prove this theorem, we can use the following helper function: $$h(x) = f(x) - \frac{f(b) - f(a)}{g(b) - g(a)} \cdot g(x)$$ The helper function is continuous and differentiable on the interval $$(a, b)$$, because $$f(x)$$ and $$g(x)$$ are continuous and differentiable on the same interval.

## Step 3: Apply Rolle's theorem

Since $$h(a) = h(b)$$, and $$h(x)$$ is continuous on the closed interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$, we can apply Rolle's theorem. By Rolle's theorem, there exists at least one $$c$$ in the open interval $$(a, b)$$ such that $$h'(c) = 0$$.

## Step 4: Compute the derivative of the helper function

Now, we compute the derivative of the helper function: $$h'(x) = f'(x) - \frac{f(b) - f(a)}{g(b) - g(a)} \cdot g'(x)$$ Since $$h'(c) = 0$$, we have: $$f'(c) - \frac{f(b) - f(a)}{g(b) - g(a)} \cdot g'(c) = 0$$

## Step 5: Isolate the desired quotient

Now, we isolate the desired quotient: $$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$ This completes the proof, and we have successfully demonstrated the existence of $$c$$ in $$(a, b)$$ satisfying the Cauchy Mean Value Theorem.

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