Problem 38 - Thomas Calculus

Question

Find all values of \(c,\) that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval. a. \(f(x)=x, \quad g(x)=x^{2}, \quad(a, b)=(-2,0)\) b. \(f(x)=x, \quad g(x)=x^{2}, \quad(a, b)\) arbitrary c. \(f(x)=x^{3} / 3-4 x, \quad g(x)=x^{2}, \quad(a, b)=(0,3)\)

Step-by-Step Solution

Answer

a) \(c = -1\); b) \(c = \frac{b^2-a^2}{2(b-a)}\); c) \(c = 1+\sqrt{5}\).

Step 1: State Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem states that for two functions \(f\) and \(g\) continuous on \([a, b]\) and differentiable on \((a, b)\), there exists at least one \(c\in(a, b)\) such that \(\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}\). This step will help us understand what we are looking for in each part.

Step 2: Apply Theorem to Part a

For part a, we have \(f(x) = x\) and \(g(x) = x^2\) on \((-2, 0)\). First, compute the derivatives: \(f'(x) = 1\) and \(g'(x) = 2x\). Then calculate \(\frac{f'(x)}{g'(x)} = \frac{1}{2x}\). Compute \(f(b) - f(a) = 0 + 2 = 2\) and \(g(b) - g(a) = 0 - 4 = -4\). Therefore, \(\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{2}{-4} = -\frac{1}{2}\). Solve \(\frac{1}{2c} = -\frac{1}{2}\) to find \(c = -1\).

Step 3: Solve Part b for Arbitrary (a, b)

For arbitrary \((a, b)\), the formula \(\frac{f'(c)}{g'(c)}=\frac{f(b) - f(a)}{g(b) - g(a)}\) becomes \(\frac{1}{2c} = \frac{b-a}{b^2-a^2}\). Solving for \(c\), multiply both sides by \(2c\) and solve: \(1 = \frac{2c(b-a)}{b^2-a^2}\), giving \(c = \frac{b^2-a^2}{2(b-a)}\).

Step 4: Apply Theorem to Part c

For part c, the functions are \(f(x) = \frac{x^3}{3} - 4x\) and \(g(x) = x^2\), on \((0, 3)\). Derivatives are \(f'(x) = x^2 - 4\) and \(g'(x) = 2x\). Compute \(\frac{f'(c)}{g'(c)} = \frac{c^2 - 4}{2c}\) and set it equal to \(\frac{f(3) - f(0)}{g(3) - g(0)}\), which is \(-\frac{1}{2}\) (after calculating). Solve \(\frac{c^2 - 4}{2c} = -\frac{1}{2}\) which yields \(c^2 = 2c + 4\), simplifying to \(c = 1 + \sqrt{5}\) or \(c = 1 - \sqrt{5}\), but valid \(c\) is \(1 + \sqrt{5}\) within \((0, 3)\).

Key Concepts

Continuity

Continuity is an essential foundation of calculus. When we say a function is continuous on an interval \[a, b\], it means that there are no breaks, jumps, or holes in the graph of the function over that interval. A continuous function smoothly connects all points within the interval without interruptions. To visualize continuity, imagine drawing a function's graph without lifting your pencil from the paper. The concept applies to both of the functions involved in Cauchy's Mean Value Theorem. Both functions must be continuous on the closed interval \[a, b\].For each subproblem in the exercise, ensuring continuity means verifying the functions behave predictably across their respective intervals. Without continuity, the theorem wouldn't hold, as the necessary conditions wouldn't be met.

Differentiability

Differentiability is closely related to continuity but delves deeper into a function's ability to have a derivative everywhere within an open interval \(a, b\). For a function to be differentiable at a point, it must also be continuous there, but a continuous function isn't necessarily differentiable. A function that is well-behaved, smooth, and has no sharp corners or cusps is considered differentiable. For Cauchy's Mean Value Theorem, requiring differentiability on the open interval \(a, b\) ensures we can compute derivatives at all points except possibly the endpoints.This requirement is crucial as derivatives form the core part of the theorem's conclusion, where we compare ratios of derivatives and differences in function values as posed in the theorem.

Derivative

The derivative represents the rate of change of a function with respect to a variable. It tells us how fast a function's value is changing at a particular point and is a fundamental tool in calculus analysis.To employ Cauchy's Mean Value Theorem, derivatives of the functions involved must be calculated accurately. For instance, in the given exercise, the derivative of \(f(x) = x\) is \(f'(x) = 1\) and for \(g(x) = x^2\), the derivative is \(g'(x) = 2x\). These derivatives are vital in computing the ratio \(\frac{f'(c)}{g'(c)}\) at some point \(c\).Working with derivatives allows us to apply the theorem not just in theorem-specific problems, but also in real-life scenarios, such as predicting trends and understanding dynamic changes in systems.

Interval Analysis

Interval analysis deals with examining functions within specific ranges of their domain, characterized by intervals. In this context, infinity is replaced with practical bounds specified by the intervals. In applying Cauchy's Mean Value Theorem, we're often inspecting closed intervals \[a, b\] and open intervals \(a, b\) to fulfill the theorem's requirements.We start with a closed interval to ensure continuity and move to the open interval for differentiability. In the solution steps, these intervals were vital to finding specific values of \(c\) that satisfy the theorem. With functions like \(f(x) = x^3 / 3 - 4x\) and \(g(x) = x^2\) from the exercise, careful interval analysis allows us to determine where these functions meet the criteria set by the theorem.Thus, interval analysis helps in localized examination of functions, validating necessary conditions like continuity and differentiability, and explicitly applying results like finding specific \(c\) values.