Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 25
Find the best possible bounds for the function. $$x+\sin x, \quad \text { for } 0 \leq x \leq 2 \pi$$
Short Answer
Expert verified
The bounds for the function are from 0 to \(2\pi\).
Step by step solution
01
Determine the Components
The function we are analyzing is \( f(x) = x + \sin x\). It is composed of the linear term \( x \) and the sine function, \( \sin x \). We need to consider how each component behaves within the interval \( [0, 2\pi] \).
02
Analyze the Linear Component
Within the interval \( [0, 2\pi] \), the linear term \( x \) increases steadily from 0 to \( 2\pi \). Hence, \( x \) takes its minimum value at 0 and its maximum at \(2\pi\).
03
Analyze the Sine Component
The function \( \sin x \) oscillates between -1 and 1. In the interval \( [0, 2\pi] \), \( \sin x \) starts at 0, reaching its maximum value of 1 at \( \frac{\pi}{2} \), and its minimum value, -1, at \( \frac{3\pi}{2} \).
04
Consider Boundary and Critical Points
To find the bounds of \( f(x) \), consider both endpoints \( x = 0 \) and \( x = 2\pi \), and any critical points within the interval where the derivative \( f'(x) = 1 + \cos x = 0 \). The critical points are where \( \cos x = -1 \), namely, \( x = \pi \).
05
Evaluate Function at Critical and Boundary Points
Calculate \( f(x) \) at these points:- \( f(0) = 0 + \sin 0 = 0 \)- \( f(2\pi) = 2\pi + \sin 2\pi = 2\pi \)- \( f(\pi) = \pi + \sin \pi = \pi \)We should also check other key values, such as at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \). Check them:- \( f(\frac{\pi}{2}) = \frac{\pi}{2} + \sin \frac{\pi}{2} = \frac{\pi}{2} + 1\)- \( f(\frac{3\pi}{2}) = \frac{3\pi}{2} + \sin \frac{3\pi}{2} = \frac{3\pi}{2} - 1 \)
06
Determine the Bounds
After calculating the values at critical and boundary points, compare them:- \( f(0) = 0 \)- \( f(2\pi) = 2\pi \)- \( f(\pi) = \pi \)- \( f(\frac{\pi}{2}) = \frac{\pi}{2} + 1 \)- \( f(\frac{3\pi}{2}) = \frac{3\pi}{2} - 1 \)The minimum value of \( f(x) \) in the given interval is 0, and the maximum value is \( 2\pi \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function analysis
Function analysis is crucial for understanding how a function behaves across a given interval. In this exercise, we are examining the function \( f(x) = x + \sin x \), which is a combination of a linear function and a trigonometric function. Function analysis involves decomposing the function into its components and studying their behavior individually and collectively over the interval \([0, 2\pi]\). This helps in identifying patterns, extremities, and the overall trend of the function.
Trigonometric functions
Trigonometric functions, such as \( \sin x \), play a vital role in calculus due to their oscillatory nature. They are defined using circles and angles, which introduces periodic patterns. In the interval \([0, 2\pi]\), \( \sin x \) starts at 0, peaks at 1 when \( x = \frac{\pi}{2} \), dips to -1 at \( x = \frac{3\pi}{2} \), and then returns to 0 when \( x = 2\pi \). Understanding this sinusoidal pattern is essential for analyzing how \( \sin x \) impacts the function \( f(x) = x + \sin x \).
Critical points
Critical points of a function occur where its derivative is zero or undefined. These points are crucial as they identify potential maxima or minima within an interval. For \( f(x) = x + \sin x \), we compute the derivative \( f'(x) = 1 + \cos x \). Setting the derivative equal to zero gives us the critical points. In this scenario, \( \cos x = -1 \) at \( x = \pi \). Assessing these points helps in understanding where significant changes in the function's behavior might happen.
Derivative
In calculus, a derivative represents the rate of change of a function with respect to a variable. For \( f(x) = x + \sin x \), the derivative \( f'(x) = 1 + \cos x \) provides information about the slope of the function at any point within its domain. When the derivative equals zero, it signals a plateau where the function reaches a potential peak or trough. Therefore, calculating and analyzing the derivative is vital in identifying key points like maxima and minima of the function.
Interval evaluation
Interval evaluation is assessing a function's behavior across a specified range of values. Here, we analyze \( f(x) = x + \sin x \) over \([0, 2\pi]\). This includes looking at endpoints and critical points. By evaluating \( f(x) \) at the interval's boundaries and calculated critical points鈥擻( x = 0, 2\pi, \frac{\pi}{2}, \pi, \text{and } \frac{3\pi}{2} \)鈥攚e determine the function's range of values. The findings show that within this interval, \( f(x) \) achieves a minimum value of 0 and a maximum value of \( 2\pi \).
