Problem 72
Question
Use the Mean Value Theorem to show that \((1+x)^{p}<1+p x\) for \(0
Step-by-Step Solution
Verified Answer
Using the Mean Value Theorem, \( (1+x)^p < 1 + px \) for \( 0 < x \) and \( p < 1 \).
1Step 1: Understand the Mean Value Theorem (MVT)
The Mean Value Theorem states that for a function \( f \) that is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a point \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
2Step 2: Define the Function
Consider the function \( f(x) = (1+x)^p \). We want to investigate this function over the interval \([0, x]\).
3Step 3: Calculate the Derivative
Find the derivative of \( f(x) = (1+x)^p \). By using the power rule, \( f'(x) = p(1+x)^{p-1} \).
4Step 4: Apply MVT to the Interval [0, x]
According to MVT, there exists a \( c \in (0, x) \) such that \( f'(c) = \frac{(1+x)^p - 1^p}{x} \). Substituting \( f'(c) \), gives us \( p(1+c)^{p-1} = \frac{(1+x)^p - 1}{x} \).
5Step 5: Analyze the Inequality
Since \( p < 1 \) and \( 0 < x \), then \((1+c)^{p-1} < 1 \). Therefore, \( p(1+c)^{p-1} < p \). Thus, \( \frac{(1+x)^p - 1}{x} < p \) which results in \((1+x)^p < 1 + px\).
6Step 6: Conclude the Inequality
We've shown that \( (1+x)^p < 1 + px \) by using the Mean Value Theorem and identifying that \( p(1+c)^{p-1} < p \) due to the condition \( p < 1 \).
Key Concepts
CalculusDifferentiationInequality AnalysisPower Rule
Calculus
Calculus is a branch of mathematics that deals with continuous change. It is divided into two main parts: differentiation, which involves finding rates of change and slopes of curves, and integration, which is concerned with calculating areas under curves and the accumulation of quantities. Calculus is essential in many fields including science, engineering, and economics because it provides a framework for modeling and solving real-world problems involving change.
In this exercise, we employ the Mean Value Theorem, a fundamental concept within calculus, to establish an inequality. The theorem helps us find a specific point in an interval where the slope of the tangent (derivative) matches the slope of the secant line connecting two points on the curve. This forms a core connection between the behavior of a function and its derivative over an interval.
In this exercise, we employ the Mean Value Theorem, a fundamental concept within calculus, to establish an inequality. The theorem helps us find a specific point in an interval where the slope of the tangent (derivative) matches the slope of the secant line connecting two points on the curve. This forms a core connection between the behavior of a function and its derivative over an interval.
Differentiation
Differentiation is the process of finding the derivative of a function, which is a key tool in calculus. The derivative represents the rate at which one quantity changes with respect to another. It is symbolically represented as \(f'(x)\), and it provides insight into the behavior and shape of graphs.
The exercise requires computing the derivative of the function \(f(x) = (1+x)^p\). By applying the power rule, we can differentiate the function, yielding \(f'(x) = p(1+x)^{p-1}\). This derivative allows us to apply the Mean Value Theorem over the interval \([0, x]\) to analyze the function’s behavior and derive an inequality. Understanding differentiation is crucial, as it allows for the detailed analysis of functions and their rates of change.
The exercise requires computing the derivative of the function \(f(x) = (1+x)^p\). By applying the power rule, we can differentiate the function, yielding \(f'(x) = p(1+x)^{p-1}\). This derivative allows us to apply the Mean Value Theorem over the interval \([0, x]\) to analyze the function’s behavior and derive an inequality. Understanding differentiation is crucial, as it allows for the detailed analysis of functions and their rates of change.
Inequality Analysis
Inequality analysis involves comparing quantities to establish the relative size or value of different mathematical expressions. In this exercise, we used the Mean Value Theorem to illustrate that \( (1+x)^p < 1 + px \). Inequalities often provide bounds or limitations for mathematical models that can be crucial for making predictions or decisions in various applications.
We explored the inequality by substituting the derivative derived in the previous section into the Mean Value Theorem. This comparison of the secant slope and tangent slope led us to establish the critical condition: \( p(1+c)^{p-1} < p \), which subsequently demonstrated that \( (1+x)^p \) is less than \( 1 + px \) given \(0
We explored the inequality by substituting the derivative derived in the previous section into the Mean Value Theorem. This comparison of the secant slope and tangent slope led us to establish the critical condition: \( p(1+c)^{p-1} < p \), which subsequently demonstrated that \( (1+x)^p \) is less than \( 1 + px \) given \(0
Power Rule
The power rule is a basic principle in calculus used for differentiating functions of the form \(x^n\). It states that the derivative of \(x^n\) is \(nx^{n-1}\). This rule makes finding derivatives straightforward when dealing with polynomial functions.
In the exercise, the power rule was applied to the function \(f(x) = (1+x)^p\), resulting in the derivative \(f'(x) = p(1+x)^{p-1}\). This derivative plays a crucial role when applying the Mean Value Theorem to determine the nature of the function within a given interval. Understanding and utilizing the power rule correctly is fundamental for anyone studying calculus, as it simplifies the process of finding derivatives for many common functions, allowing for broader analysis and application.
In the exercise, the power rule was applied to the function \(f(x) = (1+x)^p\), resulting in the derivative \(f'(x) = p(1+x)^{p-1}\). This derivative plays a crucial role when applying the Mean Value Theorem to determine the nature of the function within a given interval. Understanding and utilizing the power rule correctly is fundamental for anyone studying calculus, as it simplifies the process of finding derivatives for many common functions, allowing for broader analysis and application.
Other exercises in this chapter
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