/*! 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 21 (a) find the intervals on which ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 21

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x+\frac{1}{x} $$

Short Answer

Expert verified
The function \(f(x) = x + \frac{1}{x}\) has a first derivative \(f'(x) = 1 - \frac{1}{x^2}\). The function is increasing on the intervals \((-\infty, -1)\) and \((1, \infty)\), and decreasing on the interval \((-1, 1)\). There is a relative maximum at \((-1, -2)\) and a relative minimum at \((1, 2)\).

Step by step solution

01

Find the first derivative of the function

To determine if the function is increasing or decreasing, we need to find its first derivative. The function is given as \(f(x) = x + \frac{1}{x}\). To find the first derivative, we use the power rule for differentiation: \[ f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} \right) = \frac{d}{dx} (x) + \frac{d}{dx} \left( \frac{1}{x} \right) \] Now, differentiating each term, we get: \[ f'(x) = 1 - \frac{1}{x^2} \]
02

Find the critical points

Critical points occur where the first derivative is either equal to zero or undefined. In this case, the first derivative is: \[ f'(x) = 1 - \frac{1}{x^2} \] To find where the derivative is equal to zero, we set it to zero: \[ 1 - \frac{1}{x^2} = 0 \] Solving for x, we get: \[ x^2 = 1 \\ x = \pm 1 \] The first derivative is undefined when the denominator is zero, but in this case, the denominator is always nonzero. Therefore, there are no additional critical points.
03

Determine the intervals of increase and decrease

We found two critical points: \(x = -1\) and \(x = 1\). Now we need to determine the intervals on which the function is increasing or decreasing. To do that, we can analyze the sign of the first derivative in the intervals around the critical points. We test points in the intervals \((- \infty, -1)\), \((-1, 1)\), and \((1, \infty)\) by plugging them into \(f'(x) = 1 - \frac{1}{x^2}\): - For \(x < -1\), the derivative \(f'(x)\) is positive - For \(-1 < x < 1\), the derivative \(f'(x)\) is negative - For \(x > 1\), the derivative \(f'(x)\) is positive Thus, the function is increasing on the intervals \((-\infty, -1)\) and \((1, \infty)\), while it's decreasing on the interval \((-1, 1)\).
04

Find the relative maxima and minima

We can use the critical points and the intervals of increase and decrease to find the relative maxima and minima. Since the function changes from increasing to decreasing at \(x = -1\), there is a relative maximum at this point: \(f(-1) = -1 + \frac{1}{-1} = -2\) Therefore, there is a relative maximum at \((-1, -2)\). Similarly, since the function changes from decreasing to increasing at \(x = 1\), there is a relative minimum at this point: \(f(1) = 1 + \frac{1}{1} = 2\) Thus, there is a relative minimum at \((1, 2)\).
05

Summary

The given function is \(f(x) = x + \frac{1}{x}\). Its first derivative is \(f'(x) = 1 - \frac{1}{x^2}\). The function is increasing on the intervals \((-\infty, -1)\) and \((1, \infty)\) and decreasing on the interval \((-1, 1)\). There is a relative maximum at \((-1, -2)\) and a relative minimum at \((1, 2)\).

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.

Derivatives
In calculus, derivatives are a fundamental tool for understanding how a function changes. When we say we are differentiating a function, we're essentially finding out the rate at which the function's value changes with respect to changes in its input value.
The first derivative of a function, often denoted as \( f'(x) \), tells us how the function \( f(x) \) changes as \( x \) changes. For example, in our function \( f(x) = x + \frac{1}{x} \), the derivative is found using basic differentiation rules:
  • The derivative of \( x \) with respect to \( x \) is 1.
  • The derivative of \( \frac{1}{x} \), using the power rule, is \( -\frac{1}{x^2} \).
So, \( f'(x) = 1 - \frac{1}{x^2} \). This gives us a new function that helps determine where \( f(x) \) is increasing or decreasing.
Critical Points
Critical points are specific values of \( x \) where the derivative of a function is either zero or undefined. Identifying these points is crucial because they are potential locations of maxima, minima, or points of inflection.
For the function \( f(x) = x + \frac{1}{x} \), we found its first derivative to be \( f'(x) = 1 - \frac{1}{x^2} \). We set this equal to zero to find the critical points:
  • \( 1 - \frac{1}{x^2} = 0 \) solves to \( x = \pm 1 \).
These are our critical points as both make the derivative equal to zero. No points make the derivative undefined, because \( x^2 \) can never be zero in this context. Critical points like these guide us in evaluating the function's behavior at and near these points.
Increasing and Decreasing Intervals
Once we identify the critical points, the next step is to determine on which intervals the function is increasing or decreasing. This involves evaluating the sign of the first derivative in different intervals around the critical points.
For \( f'(x) = 1 - \frac{1}{x^2} \):
  • In the interval \((-\infty, -1)\), \( f'(x) \) is positive, indicating that \( f(x) \) is increasing.
  • In the interval \((-1, 1)\), \( f'(x) \) is negative, so \( f(x) \) is decreasing.
  • In the interval \((1, \infty)\), \( f'(x) \) is positive, indicating an increase in \( f(x) \).
These intervals help us understand where the function is "going up" or "going down," leading us to zones of potential maximum or minimum values.
Relative Maxima and Minima
Relative maxima and minima refer to the points on a graph where a function reaches a high or low value, relative to its immediate surroundings. A relative maximum happens when a function changes from increasing to decreasing, while a relative minimum occurs when it shifts from decreasing to increasing.
For the function \( f(x) = x + \frac{1}{x} \), we find:
  • At \( x = -1 \), the function changes from increasing to decreasing, giving a relative maximum at \((-1, -2)\).
  • At \( x = 1 \), the function transitions from decreasing to increasing, creating a relative minimum at \((1, 2)\).
Identifying these points is essential for understanding the complete behavior of \( f(x) \), aiding in the analysis of the function's full graph.

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

Terminal Velocity A skydiver leaps from the gondola of a hotair balloon. As she free-falls, air resistance, which is proportional to her velocity, builds up to a point at which it balances the force due to gravity. The resulting motion may be described in terms of her velocity as follows: Starting at rest (zero velocity), her velocity increases and approaches a constant velocity, called the terminal velocity. Sketch a graph of her velocity \(v\) versus time \(t\).

Use Newton's method to find the point of intersection of the graphs to four decimal places of accuracy by solving the equation \(f(x)-g(x)=0 .\) Use the initial estimate \(x_{0}\) for the \(x\) -coordinate. f(x)=\frac{1}{2} \cos x, g(x)=x, \quad x_{0}=0.5

Tracking a Submarine A submarine traveling along a path described by the equation \(y=x^{2}+1\) (both \(x\) and \(y\) are measured in miles) is being tracked by a sonobuoy (sound detector) located at the point \((3,0)\) in the figure. a. Show that the submarine is closest to the sonobuoy if \(x\) satisfies the equation \(2 x^{3}+3 x-3=0\). Hint: Minimize the square of the distance between the points \((3,0)\) and \((x, y)\). b. Use Newton's method to solve the equation \(2 x^{3}+3 x-3=0\) c. What is the distance between the submarine and the sonobuoy at the closest point of approach? I d. Plot the graph of the function \(g(x)\) giving the distance between the submarine and the sonobuoy using the viewing window \([0,1] \times[2,4]\). Then use it to verify the result that you obtained in parts (b) and (c).

Mass of a Moving Particle The mass \(m\) of a particle moving at a speed \(v\) is related to its rest mass \(m_{0}\) by the equation $$ m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$ where \(c\), a constant, is the speed of light. Show that $$ \lim _{v \rightarrow c^{-}} \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\infty $$ thus proving that the line \(v=c\) is a vertical asymptote of the graph of \(m\) versus \(v .\) Make a sketch of the graph of \(m\) as a function of \(v\).

Effect of Inflation on Salaries Mr. Gilbert's current annual salary is \(\$ 75,000\). Ten years from now, how much will he need to earn to retain his present purchasing power if the rate of inflation over that period is \(5 \%\) per year? Assume that inflation is compounded continuously.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Calculus

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.