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Chapter 4: Problem 58

Show that the given inequality holds for the given values of \(\mathrm{x}\). $$ 1 / x+\tan (1 / x)<1+\pi / 4 \text { for } x>4 / \pi $$

Short Answer

Expert verified
The given inequality holds because the combined terms approach values less than the right side for \( x> 4/\pi \).

Step by step solution

01

Analyze Given Inequality

We need to show that the inequality \( \frac{1}{x} + \tan\left(\frac{1}{x}\right) < 1 + \frac{\pi}{4} \) holds for \( x > \frac{4}{\pi} \). This requires analyzing the behavior of the function \( \frac{1}{x} + \tan\left(\frac{1}{x}\right) \) as \( x \) increases.
02

Understand Function Behavior

For \( x > \frac{4}{\pi} \), the term \( \frac{1}{x} \) becomes small, hence \( \frac{1}{x} < \frac{\pi}{4} \). The function \( \tan z \) approaches \( z \) when \( z \) (i.e., \( \frac{1}{x} \)) is small. Therefore, \( \tan\left(\frac{1}{x}\right) \approx \frac{1}{x} \), simplifying the inequality.
03

Simplify Inequality

Using \( \tan\left(\frac{1}{x}\right) \approx \frac{1}{x} \), the expression \( \frac{1}{x} + \tan\left(\frac{1}{x}\right) \approx \frac{2}{x} \). Given \( x > \frac{4}{\pi} \), \( \frac{2}{x} < \frac{\pi}{2} \), which is indeed less than \( 1 + \frac{\pi}{4} \).
04

Verify Condition

We verify that for values of \( x > \frac{4}{\pi} \), \( \frac{2}{x} < \frac{\pi}{2} \) holds: \( x > \frac{4}{\pi} \Rightarrow \frac{2}{x} < \frac{\pi}{2} \). This confirms our inequality \( \frac{1}{x} + \tan\left(\frac{1}{x}\right) < 1 + \frac{\pi}{4} \).

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Key Concepts

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

Tangent Function
The tangent function, often represented as \( \tan(z) \), is one of the fundamental trigonometric functions related to angles and triangles. For small values of \( z \), the function behaves in a nearly linear manner, which is why it can be approximated as \( \tan(z) \approx z \). This approximation is crucial when dealing with the tangent of small angles, like \( \frac{1}{x} \) when \( x \) is large. Understanding this behavior helps us analyze inequalities involving \( \tan \) more easily, especially since it simplifies complex expressions into manageable linear functions.
Simplifying Expressions
Simplifying expressions is a key step in solving calculus problems, particularly inequalities. In this exercise, we have the expression \( \frac{1}{x} + \tan\left(\frac{1}{x}\right) \). By recognizing that for small \( \frac{1}{x} \), the tangent function \( \tan\left(\frac{1}{x}\right) \approx \frac{1}{x} \), we can simplify the expression to \( \frac{2}{x} \).
  • This simplification makes it easier to analyze and solve the inequality.
  • It reduces the complexity of the original equation.
  • It highlights the mathematical behavior of the function more clearly.
Inequality Analysis
Inequality analysis involves understanding and proving the conditions under which an inequality holds. In the given problem, we start with the inequality \( \frac{1}{x} + \tan\left(\frac{1}{x}\right) < 1 + \frac{\pi}{4} \). We then simplify our expressions, resulting in \( \frac{2}{x} < \frac{\pi}{2} \). To verify this:
  • Check if \( x > \frac{4}{\pi} \) satisfies this inequality.
  • Ensure that the simplified expression embodies the stipulated condition.
By confirming these, we better understand the behavior and relationships of functions involved in inequalities.
Calculus Problem Solving
Calculus problem solving often requires an array of skills, including simplifying expressions and understanding the behavior of functions. The given exercise tests these skills by asking us to prove an inequality involving specific function behaviors and approximations. Successful calculus problem solving involves:
  • Breaking down a problem into manageable steps (e.g., analyze, simplify, verify).
  • Applying mathematical principles to simplify complex functions.
  • Using small-value approximations to solve inequalities.
These strategies aid in grasping the problem's nuances and ensuring accurate solutions.

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Most popular questions from this chapter

Find the horizontal asymptote of the graph of the function. Then sketch the graph of the function. $$ f(x)=\frac{2 x-3}{4-6 x} $$

Suppose an object is released and moves through a viscous fluid that tends to resist the motion of the object. Then the velocity \(v\) increases with time, and may approach a terminal velocity \(v_{T}\), which depends on the mass of the object and the viscosity of the fluid. If the fluid's resistance is proportional to the object's velocity, then as a function of time \(t\) the velocity is given by $$ v=v_{T}\left(1-e^{-g t / v_{T}}\right) $$ where \(g=9.8\) (meters per second per second) is the acceleration due to gravity. a. Find \(\lim _{t \rightarrow \infty} v(t)\). b. If the object is a tiny fog droplet (which is frequently on the order of \(5 \times 10^{-6}\) meters in radius), and is falling in the sky near earth, then a reasonable value for the terminal velocity is \(2.7 \times 10^{-2}\) meters per second (which is equivalent to 1 meter every 37 seconds). If the fog droplet begins falling at time \(t=0\), determine how long it takes for the velocity of the droplet to reach half of the terminal velocity.

Suppose the current \(I(t)\) flowing in an electrical circuit at time \(t\) is given by $$ I(t)=\frac{100}{1+t^{2}}+3 \sin \frac{30 t}{\pi} \text { for } t \geq 0 $$ Show that $$ \lim _{t \rightarrow \infty}\left(I(t)-3 \sin \frac{30 t}{\pi}\right)=0 $$ Thus for large values of \(t, I(t)\) is very nearly equal to \(3 \sin (30 t / \pi)\). The expression \(3 \sin (30 t / \pi)\) is called the steady-state current, and the expression \(100 /\left(1+t^{2}\right)\) is the transient current (since it is significant only for small values of \(t\) ).

Find all inflection points (if any) of the graph of the function. Then sketch the graph of the function. $$ g(x)=3 x^{4}+4 x^{3} $$

Find the given limit. $$ \lim _{x \rightarrow \infty}\left(x-\sqrt{4 x^{2}-1}\right) $$
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