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Problem 123

State and prove the Cauchy Mean Value Theorem.

Short Answer

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.
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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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Most popular questions from this chapter

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