/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 130 Let \(f(x)=\frac{\ln x}{x}\). ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 130

Let \(f(x)=\frac{\ln x}{x}\). (a) Graph \(f\) on \((0, \infty)\) and show that \(f\) is strictly decreasing on \((e, \infty)\). (b) Show that if \(e \leq AB^{A}\). (c) Use part (b) to show that \(e^{\pi}>\pi^{e}\).

Short Answer

Expert verified
The function \(f(x)\) is strictly decreasing for \(x>e\). The inequality \(A^{B}>B^{A}\) holds for \(e \leq A \pi^{e}\).

Step by step solution

01

Analyzing the Function

To visualize the problem, we need to plot \(f(x)\) and check its shape on the interval \((0, \infty)\). The analyzis of the derivative helps to clarify the view.
02

Derivation of the given Function

The derivative of \(f(x)\) is \(f'(x) = \frac{1 - \ln x}{x^2}\). Setting \(f'(x) = 0\), we find \(x = e\). This means \(f(x)\) realistically has a turning point at \(x = e\). For \(x > e\), the function is strictly decreasing.
03

Verification of the Inequality

To prove \(A^{B} > B^{A}\) for \(e \leq A < B\), we noted that \(\frac{\ln A}{A} > \frac{\ln B}{B}\), which is valid as \(A > e\) and \(B > A\). So, we can say that, \(\log_{A}{B} > \log_{B}{A}\) and \(\log_{A}{B} < 1\). Thus, \(A^{B} > B^{A}\).
04

Applying the inequality to Number Comparisons

To show that \(e^{\pi} > \pi^{e}\) we use results from Step 3. As \(e < \pi\), we know that \(A^B > B^A\) meaning, \(e^{\pi} > \pi^{e}\), which is our desired result.

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Key Concepts

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

Derivative
Understanding the derivative of a function is crucial to analyze its behavior. A derivative reflects how a function changes as its input changes. In simpler terms, it's like the speed of a function. For the given function, \(f(x) = \frac{\ln x}{x}\), the derivative \(f'(x) = \frac{1 - \ln x}{x^2}\) gives us insight into how the function increases or decreases. By setting \(f'(x) = 0\), we identified \(x = e\) as a turning point. Utilizing the derivative helps us determine that for \(x > e\), the function decreases continuously.
  • The derivative becomes negative when \(x > e\)
  • This negative derivative indicates a decrease in function
Recognizing the value of \(x = e\) as a critical point highlights where the behavior of \(f(x)\) changes, assisting in sketching the graph accurately.
Inequality
Inequalities express the relative size or order of two values. They tell us when one quantity is larger or smaller than another. In this context, we examine the inequality \(A^B > B^A\) for \(e \leq A < B\). Inequalities are crucial in comparing exponential expressions, particularly when involving variables.
  • The inequality signifies that the power \(A^B\) exceeds the power \(B^A\)
  • This happens in ranges where \(e \leq A < B\)
Using the property \(\frac{\ln A}{A} > \frac{\ln B}{B}\) helps conclude the inequality, showing how certain logs can determine the size of exponential expressions without direct calculation.
Exponential Functions
Exponential functions are functions in which a constant base is raised to a variable exponent. These functions grow rapidly and are significant in various fields such as mathematics, physics, and finance. We specifically deal with expressions involving powers in the form \(A^B\) and \(B^A\). Understanding such comparisons helps in proving inequalities like \(e^\pi > \pi^e\).
  • They exhibit quick growth depending on the exponent
  • Involves comparing sizes of expressions to understand ordering
In the given context, the property of exponential growth is used by leveraging logarithmic inequalities to compare the expressions effectively.
Decreasing Functions
A decreasing function is one where as the input increases, the output steadily decreases. With the function \(f(x) = \frac{\ln x}{x}\), we determined that it is decreasing on \((e, \infty)\). The derivative \(f'(x) < 0\) on this interval confirms this behavior. Recognizing a function as decreasing is essential for solving inequalities and other calculus problems.
  • Providing that \(f'(x) < 0\) is crucial in confirming a decreasing trend
  • Helps show the ordering required in the inequality comparison
Noticing this decreasing nature simplifies the logical deduction needed to establish and validate inequalities such as the one discussed, where exponential expressions are compared.

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Most popular questions from this chapter

Let \(f\) and \(g\) be one-to-one functions. Prove that (a) \(f \circ g\) is one-to- one and (b) \((f \circ g)^{-1}(x)=\left(g^{-1} \circ f^{-1}\right)(x)\).

Find the limit. $$ \lim _{x \rightarrow-\infty} \operatorname{csch} x $$

Find the integral. $$ \int \frac{\cosh x}{\sinh x} d x $$

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

Solve the differential equation. $$ \frac{d y}{d x}=\frac{1}{(x-1) \sqrt{-4 x^{2}+8 x-1}} $$
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