Problem 78 - University Calculus: Early Transcendentals

Question

Find all values of \(c\) that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval. a. \(f(x)=x, \quad g(x)=x^{2}\) \((a, b)=(-2,0)\) b. \(f(x)=x, \quad g(x)=x^{2}\) \((a, b)\) 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 = \((b^2-a^2)/(2(b-a))\); c) c = 1 + \sqrt{5}.

Step 1: Understand Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem states that if two functions \(f(x)\) and \(g(x)\) are continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \(c \in (a, b)\) such that \(\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}\).

Step 2: Apply Cauchy's Theorem to part (a)

For \(f(x) = x\) and \(g(x) = x^2\), first find \(f'(x) = 1\) and \(g'(x) = 2x\). Using the given interval \((-2, 0)\), calculate \(\frac{f(0) - f(-2)}{g(0) - g(-2)} = \frac{0 - (-2)}{0 - 4} = \frac{2}{-4} = -\frac{1}{2}\). We then set \(\frac{f'(c)}{g'(c)} = \frac{1}{2c} = -\frac{1}{2}\). Solving \(1 = -c\) gives \(c = -1\).

Step 3: Generalize for part (b)

On an arbitrary interval \((a, b)\), use the same derivatives \(f'(x) = 1\) and \(g'(x) = 2x\). The formula \(\frac{f(b) - f(a)}{g(b) - g(a)}\) becomes \(\frac{b-a}{b^2-a^2}\). Solving \(\frac{1}{2c} = \frac{b-a}{b^2-a^2}\) leads to \(c = \frac{b^2-a^2}{2(b-a)}\) under the assumption \(a eq b\).

Step 4: Apply Cauchy's Theorem to part (c)

For \(f(x) = \frac{x^3}{3} - 4x\) and \(g(x) = x^2\), calculate \(f'(x) = x^2 - 4\) and \(g'(x) = 2x\). With \((a, b) = (0, 3)\), find \(\frac{f(3) - f(0)}{g(3) - g(0)} = \frac{9 - 0}{9 - 0} = 1\). We have \(\frac{f'(c)}{g'(c)} = \frac{c^2 - 4}{2c} = 1\). Solving \(c^2 - 4 = 2c\) yields \(c^2 - 2c - 4 = 0\). Solving this quadratic equation, \(c = 1 + \sqrt{5}\) or \(c = 1 - \sqrt{5}\), but only \(c = 1 + \sqrt{5}\) is valid within \((0, 3)\).

Key Concepts

Continuous Functions

Continuous functions are essential in calculus as they ensure that a function's output changes smoothly without any jumps or breaks. A function is continuous if you can draw its graph without lifting your pencil from the paper. In mathematical terms, a function \( f(x) \) is continuous at a point \( x = c \) if the following conditions are met:

• \( f(c) \) must be defined.
• The limit of \( f(x) \) as \( x \) approaches \( c \) must exist.
• The limit of \( f(x) \) as \( x \) approaches \( c \) must equal \( f(c) \).

These conditions ensure that as you get infinitely close to \( c \) from either direction, the function \( f(x) \) consistently approaches the same value. In Cauchy's Mean Value Theorem, continuity on a closed interval \([a, b]\) guarantees that both functions behave predictably at the boundaries and that the behaviors within the interval can be analyzed reliably.

Differentiable Functions

Differentiable functions are those that have a derivative at every point in their domain. The derivative of a function \( f(x) \) at a point \( x = c \) represents the slope of the tangent to the function's graph at that point. For a function to be differentiable at a point, it must be:

• Continuous at that point.
• Smooth, meaning it doesn't have sharp corners or cusps where the graph folds over.

These conditions suggest that differentiability is a stronger condition than continuity. Cauchy's Mean Value Theorem requires that the given functions be differentiable on the open interval \((a, b)\) because this smoothness allows us to apply the theorem and find a specific \( c \) within the interval where the theorem holds true.

Derivatives

Derivatives are a core concept in calculus, representing the rate at which a function's value changes. The derivative of a function \( f(x) \), denoted as \( f'(x) \), gives the slope of the function's graph at any point \( x \). Key points about derivatives include:

• The derivative is a limit: \( f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \).
• Common derivatives to know are: \( (x^n)' = nx^{n-1} \) and \( (c)' = 0 \) for a constant \( c \).

Understanding derivatives is crucial when applying Cauchy's Mean Value Theorem. You find the derivatives of the functions involved, and the theorem provides a specific point \( c \) where their ratio equals the ratio of their overall change across the interval. This process transforms into solving an equation to find the valid \( c \).

Quadratic Equation

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). They're called 'quadratic' because they include the second power of the unknown variable. Solving quadratic equations is a fundamental skill in algebra and is particularly useful in calculus when dealing with problems like those in Cauchy's Mean Value Theorem. The methods to solve quadratic equations include:

• Factoring: If you can express \( ax^2 + bx + c \) as a product of two binomials, you can directly find the solutions.
• Quadratic Formula: The solutions for \( ax^2 + bx + c = 0 \) are given by \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).
• Completing the Square: Transforming the quadratic into a perfect square trinomial enables finding the roots.

In part (c) of the original exercise, you used these techniques to solve \( c^2 - 2c - 4 = 0 \) for the correct value of \( c \). The ability to solve these efficiently helps identify the specific \( c \) satisfying Cauchy's theorem.