/*! 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 5 Find all critical numbers of the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 5

Find all critical numbers of the given function. $$ g(x)=x+1 / x $$

Short Answer

Expert verified
The critical numbers are \( x = 1 \) and \( x = -1 \).

Step by step solution

01

Find the Derivative

To find the critical numbers, we first need to determine the derivative of the function. The given function is:\[ g(x) = x + \frac{1}{x} \]Using the power rule and the derivative of a reciprocal, the derivative is:\[ g'(x) = 1 - \frac{1}{x^2} \]
02

Set the Derivative to Zero

Set the derivative equal to zero to find the critical points:\[ 1 - \frac{1}{x^2} = 0 \]Solve for \(x\):\[ 1 = \frac{1}{x^2} \]This means \( x^2 = 1 \).
03

Solve for x

Solve the equation \( x^2 = 1 \) to find the values of \(x\):\[ x = \pm 1 \]Thus, the potential critical numbers are \( x = 1 \) and \( x = -1 \).
04

Determine the Domain of the Original Function

Since the original function \( g(x) = x + \frac{1}{x} \) includes a term \( \frac{1}{x} \), the function is not defined for \( x = 0 \).Thus, \( x = 0 \) is excluded from the critical numbers.
05

Identify the Valid Critical Numbers

The critical numbers are the values from Step 3 that lie in the domain of \( g(x) \), which excludes \( x = 0 \). Thus, the critical numbers of the function are \( x = 1 \) and \( x = -1 \).

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.

The Derivative: Finding the Rate of Change
To find critical numbers of a function, determining the derivative is a crucial first step. The derivative measures how a function changes at each point. For the function \( g(x) = x + \frac{1}{x} \), finding the derivative highlights where changes stop or turn around.

Calculating derivatives involves different methods depending on the function. In this exercise, knowing how to differentiate a sum of functions, like \( x \) and \( \frac{1}{x} \), is essential.
  • The derivative of \( x \) is straightforward, resulting in 1 because each unit increase in \( x \) results in a one-unit increase in \( g(x) \).
  • For the term \( \frac{1}{x} \), its derivative relies on special derivative formulas, like the reciprocal derivative, which we will cover.
Practicing derivative calculations enhances your ability to determine critical numbers, helping you understand function behavior.
Unraveling the Power Rule
The power rule is a fundamental tool for taking derivatives, widely applicable when a function includes terms like \( x^n \). The rule simplifies finding derivatives for any power of \( x \).

To apply the power rule:
  • Multiply the coefficient by the power.
  • Reduce the power by one.
For example, if we had \( x^1 \) in our function, the power rule confirms its derivative is 1.

In this exercise, differentiating \( x \) as part of \( g(x) = x + \frac{1}{x} \) is a direct application of the power rule, showcasing its simplicity and effectiveness.
Exploring the Function Domain
Understanding the domain of a function is crucial when identifying critical numbers. The domain is the set of all possible inputs \( x \) for a function, affecting its critical points.

For \( g(x) = x + \frac{1}{x} \), notice the term \( \frac{1}{x} \). Here, \( x = 0 \) is not allowed because division by zero is undefined.

As a result, the domain of \( g(x) \) excludes \( x = 0 \). When considering critical numbers, it's important to only include those within the domain. Thus, even if solving for critical numbers gives us potential values, they must be checked against the function's domain.
Understanding Reciprocal Derivatives
The reciprocal derivative refers to the process of deriving a function like \( \frac{1}{x} \). This is essential when the function involves inverse or reciprocal terms.

For \( \frac{1}{x} \):
  • Recognize it can be expressed as \( x^{-1} \).
  • Apply the power rule: start by bringing down the power (e.g., -1) and reduce the power by one, leading to \(-x^{-2}\).
Hence, the derivative of \( \frac{1}{x} \) becomes \( -\frac{1}{x^2} \).

This derivative tells us how the reciprocal function鈥檚 rate of change behaves, key in identifying critical numbers and overall understanding the role of reciprocal functions in calculus.

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

Recall that a function \(f\) is even if \(f(-x)=f(x)\) for all \(x\), and \(f\) is odd if \(f(-x)=-f(x)\) for all \(x\). a. If \(f\) is even and its graph is concave upward on \((0, \infty)\), what is the concavity on \((-\infty, 0)\) ? b. If \(f\) is odd and its graph is concave upward on \((0, \infty)\), what is the concavity on \((-\infty, 0)\) ?

Find the given limit. $$ \lim _{x \rightarrow-\infty} e^{-1 / x} $$

a. Let \(g\) be continuous on \([a, b]\) and differentiable on \((a, b)\). Assume that \(g(x)=0\) for two values of \(x\) in \((a, b)\). Show that there is at least one value \(c\) in \((a, b)\) such that \(g^{\prime}(c)=0 .\) b. Let \(g(x)=x^{4}-20 x^{3}-25 x^{2}-x+1\). Use (a) to show that there is some \(c\) in \([-1,1]\) such that \(4 c^{3}-60 c^{2}-\) \(50 c-1=0 .\)

Show that if \(f^{*}\) exists and is continuous on an open interval containing \(c\) and if \(f\) has an inflection point at (c, \(f(c))\), then \(f^{\prime \prime}(c)=0\)

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.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

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.