/*! 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 28 In Exercises \(23-28\) , find (a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 28

In Exercises \(23-28\) , find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. $$g(x)=2 x+\cos x$$

Short Answer

Expert verified
The function \(g(x) = 2x + \cos (x)\) has no local extrema, is increasing over the interval \((- \infty, \infty)\), and doesn't decrease at any interval.

Step by step solution

01

Obtain the derivative

Differentiate the function \(g(x)\). The derivative of \(g(x) = 2x + \cos (x)\) with respect to \(x\) is given by \(g'(x) = 2 - \sin(x)\).
02

Find the critical points

Critical points occur in a function when its derivative is either zero or undefined. So, solve \(g'(x) = 0\) for \(x\). \nThis gives: \n\(2 - \sin(x) = 0\). \nHence, \(\sin(x) = 2\). \nHowever, the sine function outputs values between -1 and 1. So, there are no solutions to this equation; implying there are no critical points. Hence, there are no local extrema.
03

Find the intervals of increasing and decreasing

Given that there are no critical points, meaning there are no turning points or horizontal tangents, the function \(g(x) = 2x + \cos (x)\) will be either decreasing or increasing across its entire domain. Since \(g'(x) = 2 - \sin(x)\) is always greater than zero for all \(x\) as its minimum value is \(2 - 1 = 1\), the function \(g(x) = 2x + \cos (x)\) is increasing over the interval \((- \infty, \infty)\).
04

Conclusion

Therefore, the function \(g(x) = 2x + \cos (x)\) has no local extrema, always increasing over the interval \((- \infty, \infty)\), and doesn't decrease at any interval.

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.

Derivative
Understanding derivatives is fundamental in calculus as they describe the rate at which a function is changing at any given point. For the function \(g(x) = 2x + \cos x\), we found the derivative, \(g'(x) = 2 - \sin x\). This derivative tells us how the function \(g(x)\) behaves in terms of changes.
Here are some key things to remember about derivatives:
  • The derivative, \(g'(x)\), gives the slope of the tangent line to the function at any point \(x\).
  • If \(g'(x) > 0\), the function is increasing; if \(g'(x) < 0\), the function is decreasing.
  • For our function, since \(2 - \sin x\) is always positive, \(g(x)\) is always increasing. Even though we always calculate derivatives to understand changes, sometimes it's more about ensuring there are no changes in direction, as we see here.
This concept is critical as it acts as a tool to explore and predict how a function behaves, and helps in identifying intervals of increase or decrease. This leads us directly into discussing critical points.
Critical Points
Critical points arise where the derivative of a function is zero or undefined, indicating potential places where the function might change its direction. In our exercise, we solved \(g'(x) = 0\) to identify the critical points.
When we calculated, we arrived at \(\sin(x) = 2\). However, this equation has no solutions since the sine function only gives output values between -1 and 1. As a result, no critical points exist for \(g(x) = 2x + \cos x\).
  • Critical points help in locating local maxima or minima, known as local extrema.
  • In this case, without any roots from \(\sin(x) = 2\), there are no critical points that could indicate a change of direction in the function.
  • Thus, \(g(x)\) doesn't have any local extrema, simplifying our analysis of increasing and decreasing intervals.
Not every function will have critical points, as we see in this example, but looking for them is always an essential step in analysis.
Increasing and Decreasing Intervals
Finding increasing and decreasing intervals involves understanding how the signs of derivatives affect the function over its domain. Since the derivative \(g'(x) = 2 - \sin x\) is greater than zero for all \(x\), this confirms that the function \(g(x) = 2x + \cos x\) is increasing everywhere.
Here's how we understand these intervals:
  • If \(g'(x) > 0\), the function is increasing on that interval, meaning as \(x\) increases, \(g(x)\) also increases.
  • If \(g'(x) < 0\), the function is decreasing on that interval, implying that as \(x\) increases, \(g(x)\) decreases.
  • For our particular function, \(g'(x)\) never becomes negative, hence \(g(x)\) shows an increasing trend over the entire domain \((-\infty, \infty)\).
The absence of critical points simplified our analysis since there were no potential sections of the graph to suggest downward slopes or peaks throughout its domain. Effectively, this uniform behavior makes the function simple to describe without needing to account for complex oscillations.

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

cost, Revenue, and Profit A company can manufacture \(x\) items at a cost of \(c(x)\) dollars, a sales revenue of \(r(x)\) dollars and a profit of \(p(x)=r(x)-c(x)\) dollars (all amounts in thousands). Find \(d c / d t, d r / d t,\) and \(d p / d t\) for the following values of \(x\) and \(d x / d t\) (a) \(r(x)=9 x, \quad c(x)=x^{3}-6 x^{2}+15 x\) and \(d x / d t=0.1\) when \(x=2 .\) (b) \(r(x)=70 x, \quad c(x)=x^{3}-6 x^{2}+45 / x\) and \(d x / d t=0.05\) when \(x=1.5\)

Tolerance (a) About how accurately must the interior diameter of a 10 -m high cylindrical storage tank be measured to calculate the tank's volume to within 1\(\%\) of its true value? (b) About how accurately must the tank's exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within 5\(\%\) of the true amount?

How We Cough When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the question of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v(\) in \(\mathrm{cm} / \mathrm{sec})\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in \(\mathrm{cm}\) and \(c\) is a positive constant whose value depends in part on the length of the trachea. (a) Show that \(v\) is greatest when \(r=(2 / 3) r_{0},\) that is, when the trachea is about 33\(\%\) contracted. The remarkable fact is that \(X\) -ray photographs confirm that the trachea contracts about this much during a cough. (b) Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see to the claim that \(v\) is a maximum when \(r=(2 / 3) r_{0}\) .

Multiple Choice If \(y=\tan x, x=\pi,\) and \(d x=0.5,\) what does \(d y\) equal? \(\begin{array}{lll}{\text { (A) }-0.25} & {\text { (B) }-0.5} & {\text { (C) } 0} & {\text { (D) } 0.5}\end{array}\) (E) 0.25

Multiple Choice What is the maximum area of a right triangle with hypotenuse 10? \(\begin{array}{llll}{\text { (A) } 24} & {\text { (B) } 25} & {\text { (C) } 25 \sqrt{2}} & {\text { (D) } 48} & {\text { (E) } 50}\end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

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.