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Magazine Chapter 7: Problem 78
Find all values of \(c\) that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval. $$ \begin{array}{ll}{\text { a. } f(x)=x,} & {g(x)=x^{2}, \quad(a, b)=(-2,0)} \\\ {\text { b. } f(x)=x,} & {g(x)=x^{2}, \quad(a, b) \text { arbitrary }} \\\ {\text { c. } f(x)=x^{3} / 3-4 x,} & {g(x)=x^{2}, \quad(a, b)=(0,3)}\end{array} $$
Short Answer
Expert verified
(a) \(c = -1\), (b) \(c = \frac{b+a}{2}\), (c) \(c = 1.236\).
Step by step solution
01
Recall Cauchy's Mean Value Theorem
Cauchy's Mean Value Theorem states that 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 \[ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}. \].
02
Apply Cauchy's Theorem to Part (a)
For \( f(x) = x \) and \( g(x) = x^2 \) on \((a, b) = (-2,0)\), calculate the derivatives: \( f'(x) = 1 \) and \( g'(x) = 2x \). Compute \( f(b) - f(a) = 0 - (-2) = 2 \) and \( g(b) - g(a) = 0^2 - (-2)^2 = -4 \). Substitute into the theorem's equation: \( \frac{1}{2c} = \frac{2}{-4} = -\frac{1}{2} \). Solving \( 2c = -2 \), we get \( c = -1 \).
03
Apply Cauchy's Theorem to Part (b)
For \( f(x) = x \) and \( g(x) = x^2 \) with arbitrary \((a, b)\), derivatives are \( f'(x) = 1 \) and \( g'(x) = 2x \). Compute \( f(b) - f(a) = b - a \) and \( g(b) - g(a) = b^2 - a^2 \). This simplifies to \( \frac{b-a}{b^2-a^2} = \frac{1}{2c} \) or \( 2c(b-a) = b^2-a^2 \). Recognizing the difference of squares, rewrite as \( 2c = b+a \), so \( c = \frac{b+a}{2} \).
04
Verify Step 3 with an arbitrary interval example
For \( (a, b) = (1, 3) \), checking exemplification, we have \( c = \frac{1+3}{2} = 2 \). This verification demonstrates validity.
05
Apply Cauchy's Theorem to Part (c)
For \( f(x) = \frac{x^3}{3} - 4x \) and \( g(x) = x^2 \) on \((0, 3)\), calculate derivatives \( f'(x) = x^2 - 4 \) and \( g'(x) = 2x \). Compute \( f(b) - f(a) = \left(\frac{3^3}{3} - 4\times3\right) - 0 = -9 \) and \( g(b) - g(a) = 3^2 - 0 = 9 \). Substitute into the theorem's equation: \( \frac{c^2 - 4}{2c} = -1 \). Solving \( c^2 - 4 = -2c \), rearrange to \( c^2 + 2c - 4 = 0 \). Using the quadratic formula, \( c = \frac{-2 \pm \sqrt{20}}{2} \), simplify to \( c = -1 \pm \sqrt{5} \).
06
Select valid solution in interval for Part (c)
The solutions \( c = -1 \pm \sqrt{5} \) simplify to approximately \( c = 1.236 \) and \( c = -3.236 \). Only \( c = 1.236 \) lies in the interval \((0, 3)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Calculus
Calculus is the branch of mathematics that deals with rates of change and the accumulation of quantities. It is essential for understanding many concepts in mathematics, physics, engineering, and beyond. Calculus is divided into two major branches: differential calculus and integral calculus.
- Differential Calculus focuses on the concept of a derivative, which represents the rate at which a quantity changes.
- Integral Calculus involves the concept of an integral, which accumulates quantities over an interval.
Derivatives
Derivatives are central to differential calculus. They provide information about the rate of change of a function with respect to its input. More formally, the derivative of a function at a point is the slope of the tangent to the function's graph at that point.
- The derivative of a function, denoted as \( f'(x) \), gives the rate of change of the function \( f \) at any point \( x \).
- In practical terms, if you know how a function changes, you can predict future values, optimize functions, and solve real-world problems.
Continuous Functions
Continuous functions are foundational in calculus and analysis. A function is continuous if, roughly speaking, you can draw it without lifting your pencil. At any point within the interval, there should be no jumps, breaks, or holes in the graph of the function.
- A function \( f \) is continuous on an interval \([a, b]\) if it is continuous at every point in that interval.
- Continuity ensures smoothness and predictability, making it possible to apply theorems like Cauchy's Mean Value Theorem.
Differentiability
Differentiability is a condition that extends the idea of continuity. A function is differentiable if it has a derivative at every point in its domain.
- If a function \( f \) is differentiable at a point \( x \), then it is also continuous at that point.
- However, not all continuous functions are differentiable. For example, a graph with a sharp corner is continuous but not differentiable at that point.
Interval Analysis
Interval analysis deals with the examination of mathematical functions over specific intervals. Reviewing how functions behave along these intervals is crucial in calculus, particularly for verifying conditions of various theorems, including Cauchy's Mean Value Theorem.
- An interval \((a, b)\) is a set of numbers lying between two endpoints, \(a\) and \(b\), where \(a < b\).
- Closed intervals include the endpoints \([a, b]\), while open intervals, such as \((a, b)\), do not include them.
- The behavior and properties of functions over these intervals are essential for solving problems involving maxima, minima, and approximation of quantities.
