Chapter 4: Problem 78
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. \(f(x)=x^{3} / 3-4 x, \quad g(x)=x^{2}, \quad(a, b)=(0,3)\)
Short Answer
Expert verified
a) c = -1; b) c = \((b^2-a^2)/(2(b-a))\); c) c = 1 + \sqrt{5}.
Step by step solution
01
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 \(\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}\).
02
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 \(c = -1\).
03
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 \(a eq b\).
04
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 \((0, 3)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( f(c) \).
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.
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 \( c \).
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.