/*! 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 64 Find the relative extrema in the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 64

Find the relative extrema in the interval \(0

Short Answer

Expert verified
The relative maximum is at \( x = \frac{5\pi}{6} \) and the relative minimum is at \( x = \frac{7\pi}{6} \).

Step by step solution

01

Find the Derivative of the Function

To find the relative extrema, start by finding the first derivative of the function, which will help us locate the critical points. The function given is \[ f(x) = \sqrt{3} x + 2 \sin x \].The derivative of \( \sqrt{3} x \) is \( \sqrt{3} \), and the derivative of \( 2 \sin x \) is \( 2 \cos x \). Thus, the derivative of the function is:\[ f'(x) = \sqrt{3} + 2 \cos x \].
02

Determine Critical Points

Critical points occur where the derivative is zero or undefined. Set the derivative equal to zero to find critical points:\[ \sqrt{3} + 2 \cos x = 0 \].Solve for \( \cos x \):\[ 2 \cos x = - \sqrt{3} \]\[ \cos x = -\frac{\sqrt{3}}{2} \].The cosine of \( -\frac{\sqrt{3}}{2} \) corresponds to angles where \( x = \frac{5\pi}{6} \) and \( x = \frac{7\pi}{6} \) within the interval \( 0 < x < 2\pi \).
03

Evaluate the Second Derivative

To determine the nature of the critical points (whether each is a relative maximum or minimum), evaluate the second derivative:\[ f''(x) = -2 \sin x \].Evaluate the second derivative at each critical point:- At \( x = \frac{5\pi}{6} \), \( \sin \left(\frac{5\pi}{6}\right) = \frac{1}{2} \) so \( f''\left(\frac{5\pi}{6}\right) = -1 \) (negative value).- At \( x = \frac{7\pi}{6} \), \( \sin \left(\frac{7\pi}{6}\right) = -\frac{1}{2} \) so \( f''\left(\frac{7\pi}{6}\right) = 1 \) (positive value).Therefore, \( x = \frac{5\pi}{6} \) is a relative maximum, and \( x = \frac{7\pi}{6} \) is a relative minimum.
04

Confirm Results with Graphing Utility

Use a graphing utility to plot \( f(x) = \sqrt{3} x + 2 \sin x \) over the interval \( 0 < x < 2\pi \). You should observe that the graph has a peak at \( x = \frac{5\pi}{6} \) and a trough at \( x = \frac{7\pi}{6} \), which confirms the presence of a relative maximum and minimum, respectively, at these points.

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.

Relative Extrema
In calculus, knowing how to find relative extrema is extremely useful for understanding the nature of functions. Relative extrema refer to the peaks and valleys, points where a function either reaches a local maximum or a local minimum. These are important because they depict where a function increases to its highest point or decreases to its lowest point within a particular interval. In the context of the function \[ f(x) = \sqrt{3} x + 2 \sin x \]relative extrema occur in the interval \(0 < x < 2\pi \). It鈥檚 where the graph changes direction. Understanding these points helps paint a complete picture of the function's behavior.
Derivative
The derivative gives us crucial insight into the behavior of a function. It represents the rate at which the function's value changes as the input changes. For the function \[ f(x) = \sqrt{3} x + 2 \sin x \]the derivative, found through differentiation, is \[ f'(x) = \sqrt{3} + 2 \cos x \]. Finding this first derivative is the essential first step in determining the function鈥檚 critical points. Each derivative computation brings you closer to understanding where a function increases, decreases, or plateaus. This makes derivatives a powerful tool in finding and classifying relative extrema, since a zero derivative means a possible point of extreme.
Critical Points
Critical points are where the fun begins鈥攍iterally. They occur where the derivative of a function is zero or undefined. Finding these points \[ \sqrt{3} + 2 \cos x = 0 \] helps show us potential candidates for relative extrema. Solving this equation gives \[ \cos x = -\frac{\sqrt{3}}{2} \], which corresponds to specific angles in the given interval. In our case, this happens at \( x = \frac{5\pi}{6} \) and \( x = \frac{7\pi}{6} \). These angles give the critical points where further testing can reveal if they reach the peaks or troughs of our function鈥檚 graph.
Second Derivative Test
The second derivative test helps determine the nature of the critical points. It tells you whether each critical point is a relative maximum or minimum. By finding the second derivative\[ f''(x) = -2 \sin x \], and evaluating it at each critical point, we check the concavity of the function there. - At \( x = \frac{5\pi}{6} \), if \( f''(x) = -1 \), concavity is downward, indicating a relative maximum.- At \( x = \frac{7\pi}{6} \), if \( f''(x) = 1 \), concavity is upward, indicating a relative minimum.This approach confirms the nature of these points and solidifies understanding, showing how the second derivative makes identifying features of the function easy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Apply Newton's Method to \(f(x)=x^{2}+1\) with a starting value of \(x_{1}=0.5,\) and determine if the values of \(x_{2}, \ldots, x_{10}\) appear to converge. (b) Explain what is happening.

If \(f\) is a periodic function, then the locations of all absolute extrema on the interval \((-\infty,+\infty)\) can be obtained by finding the locations of the absolute extrema for one period and using the periodicity to locate the rest. Use this idea in these exercises to find the absolute maximum and minimum values of the function, and state the \(x\) -values at which they occur. \(f(x)=3 \cos \frac{x}{3}+2 \cos \frac{x}{2}\)

The accompanying figure shows the path of a fly whose equations of motion are $$x=\frac{\cos t}{2+\sin t}, \quad y=3+\sin (2 t)-2 \sin ^{2} t \quad(0 \leq t \leq 2 \pi)$$ (a) How high and low does it fly? (b) How far left and right of the origin does it fly?

Writing A speedometer on a bicycle calculates the bicycle's speed by measuring the time per rotation for one of the bicycle's wheels. Explain how this measurement can be used to calculate an average velocity for the bicycle, and discuss how well it approximates the instantaneous velocity for the bicycle.

(a) Use the Mean-Value Theorem to show that if \(f\) is differentiable on an open interval, and if \(\left|f^{\prime}(x)\right| \geq M\) for all values of \(x\) in the interval, then $$ |f(x)-f(y)| \geq M|x-y| $$ for all values of \(x\) and \(y\) in the interval. (b) Use the result in part (a) to show that $$ |\tan x-\tan y| \geq|x-y| $$ for all values of \(x\) and \(y\) in the interval \((-\pi / 2, \pi / 2)\) (c) Use the result in part (b) to show that $$ |\tan x+\tan y| \geq|x+y| $$ for all values of \(x\) and \(y\) in the interval \((-\pi / 2, \pi / 2)\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.