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Chapter 6: Problem 7

Use the Mean Value Theorem to prove that \((x-1) / x<\ln x1\). [Hint: Use the fact that \(D \ln x=1 / x\) for \(x>0 .\)

Short Answer

Expert verified
Therefore, by applying the Mean Value Theorem on the function \(ln(x)\) for the interval (1,x), and proving both sides of the inequality separately, we have proved that \((x-1) / x<\ln x1\).

Step by step solution

01

Apply Mean Value Theorem on ln(x) from 1 to x

Let's consider f(x) to be \(\ln x\) and apply MVT on [1,x] for \(x>1\). According to MVT, there exists a c in the interval (1, x) such that \(f'(c) = (f(x) - f(1)) / (x - 1)\). So, putting f'(c) and f(x) in the equation, we get \((\ln x - \ln 1) / (x - 1) = 1/c\). Simplifying this, we get \((\ln x) / (x - 1) = 1/c\).
02

Prove the left side of the inequality

Since we're considering \(x > 1\) in this exercise, we can easily see that \(c < x\). Therefore, by principle of reciprocal, \(1/c > 1/x\). Substituting 1/c in the equation from the Step 1, we arrive at the left side of our inequality: \((\ln x) / (x - 1) = 1/c \( which gives us \(\(x - 1) / x < \ln x.\)
03

Prove the right side of the inequality

Taking the limit as x approaches 1 in the inequality obtained in step 2, we get \(\lim_{x \to 1} (x - 1)/x = 0\). On the other hand, \(\lim_{x \to 1} ln(x) = 0\) as well since ln(1) equals 0. Therefore, we have \((x - 1) / x 鈮� ln(x)\) for \(x > 1\). This proves the right side of the inequality.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Functions
Logarithmic functions, commonly written as \( ext{ln}(x) \), are the inverse of exponential functions. If you think about pewnumbers, logarithms measure the time or number of steps needed to reach a certain level of growth. In mathematical terms, for a natural logarithm, if \( y = ext{ln}(x) \), then \( e^y = x \). This connection is crucial in various fields like calculus, economics, and statistics.
  • Multiplicative Change: Logarithms turn multiplication into addition, simplifying the calculation of products and quotients.
  • ln(x) Curve: The function rises slowly and is only defined for \(x > 0 \). Its fundamental property \( ext{ln}(1) = 0 \) indicates no change at \(x = 1 \).
  • Slope and Rates: The slope of \( ext{ln}(x) \) helps illustrate how quickly the exponential function grows, acting as a bridge to calculus concepts.
Derivatives of Functions
The derivative of a function gives us the rate at which one quantity changes with respect to another. For logarithmic functions like \( ext{ln}(x) \), the derivative is relatively simple: \( rac{d}{dx} ext{ln}(x) = \frac{1}{x} \).
This relationship is foundational in calculus. The derivative tells us the slope of the tangent line to the curve of \( ext{ln}(x) \). Think of it as a snapshot of the function's rate of change at any given point.
  • Tangent Line: At any point \( (c, ext{ln}(c)) \), the tangent line's slope is \( \frac{1}{c} \), highlighting the function's incremental change.
  • Mean Value Theorem: This theorem relies on derivatives to determine specific points where the function's average rate of change matches its instantaneous rate. It's pivotal in proving inequalities, like our exercise example.
  • Smoothness: Since \( ext{ln}(x) \) is differentiable for all \(x > 0\), it ensures the function's graph is smooth and continuous in this region.
Inequalities in Real Analysis
Inequalities in real analysis help us understand bounds and relationships between different expressions. They act as basic tools to establish parameters and limitations for functions. In the context of our exercise, we used inequalities to express the relationship within \( \frac{x-1}{x} < \text{ln}(x) < x-1 \) for \( x > 1 \).
  • Bounding Functions: Inequalities provide limits to functions, allowing us to understand how they behave in certain intervals.
  • Analyzing Behavior: By setting up inequalities, we assess and compare different values of functions, like how \( \text{ln}(x) \) grows between \( \frac{x-1}{x} \) and \(x-1 \).
  • Approximation: These inequalities often serve to approximate functionalities or behaviors, creating better predictions and understanding.
While inequalities might seem like rigid rules, they are flexible tools in analyzing functions. They capture the essence of function behavior such as growth, boundedness, and convergence, especially when paired with the Mean Value Theorem and derivatives.

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Most popular questions from this chapter

Suppose that \(I \subseteq \mathbb{R}\) is an open interval and that \(f^{\prime \prime}(x) \geq 0\) for all \(x \in I\). If \(c \in I\), show that the part of the graph of \(f\) on \(I\) is never below the tangent line to the graph at \((c, f(c))\).

Let \(f, g\) be differentiable on \(\mathbb{R}\) and suppose that \(f(0)=g(0)\) and \(f^{\prime}(x) \leq g^{\prime}(x)\) for all \(x \geq 0\). Show that \(f(x) \leq g(x)\) for all \(x \geq 0\)

Let \(f\) be differentiable on \((0, \infty)\) and suppose that \(\lim _{t \rightarrow \infty}\left(f(x)+f^{\prime}(x)\right)=L\). Show that \(\lim _{x \rightarrow \infty} f(x)=L\) and \(\lim _{x \rightarrow \infty} f^{\prime}(x)=0 .\left[\right.\) Hint \(\left.: f(x)=e^{x} f(x) / e^{x} .\right]\)

The function \(f(x)=8 x^{3}-8 x^{2}+1\) has two zeros in \([0,1]\). Approximate them, using Newton's Method, with the starting points (a) \(x_{1}:=\frac{1}{8}\), (b) \(x_{1}:=\frac{1}{4}\). Explain what happens.

Let \(I \subseteq \mathbb{R}\) be an interval and let \(c \in l\). Suppose that \(f\) and \(g\) are defined on \(I\) and that the derivatives \(f^{(n)}, g^{(n)}\) exist and are continuous on \(I .\) If \(f^{(k)}(c)=0\) and \(g^{(k)}(c)=0\) for \(k=0,1, \cdots, n-1\), but \(g^{(n)}(c) \neq 0\), show that $$\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\frac{f^{(n)}(c)}{g^{(n)}(c)} .$$
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