80. Find all values of c that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.
a. f(x) = x, g(x) = x², (a, b) = (-2, 0)

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Recall the statement of Cauchy's Mean Value Theorem: If functions \(f\) and \(g\) are continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \(c \in (a, b)\) such that \[\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}.\]

Verify that \(f(x) = x\) and \(g(x) = x^2\) are continuous on \([-2, 0]\) and differentiable on \((-2, 0)\), which they are since they are polynomials.

Calculate the difference quotient on the right side: \[\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f(0) - f(-2)}{g(0) - g(-2)} = \frac{0 - (-2)}{0^2 - (-2)^2} = \frac{2}{0 - 4} = \frac{2}{-4} = -\frac{1}{2}.\]

Find the derivatives: \[f'(x) = 1, \quad g'(x) = 2x.\]

Set up the equation from Cauchy's Mean Value Theorem: \[\frac{f'(c)}{g'(c)} = \frac{1}{2c} = -\frac{1}{2}.\] Solve for \(c\) in the interval \((-2, 0)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem generalizes the Mean Value Theorem by relating two functions f and g that are continuous on [a, b] and differentiable on (a, b). It guarantees the existence of a point c in (a, b) where the ratio of their derivatives equals the ratio of their increments: (f'(c)/g'(c)) = (f(b)-f(a))/(g(b)-g(a)).

Differentiability and Continuity Conditions

For Cauchy's Mean Value Theorem to apply, both functions must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). These conditions ensure the existence of the point c and the validity of the theorem's conclusion.

Computing Derivatives and Evaluating at c

To find the value(s) of c, compute the derivatives f'(x) and g'(x), then set their ratio equal to the ratio of the function increments over [a, b]. Solving this equation for c within the interval (a, b) yields the required value(s) satisfying the theorem.

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