/*! 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 In Problems , find c such that \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 21

In Problems , find c such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) $$ f(x)=-x^{2} $$

Short Answer

Expert verified
The critical point is at \( c = 0 \) and \( f(x) \) has a local maximum there.

Step by step solution

01

Find the first derivative of f(x)

To find the critical points of the function, we need to calculate its first derivative. Given \[ f(x) = -x^2 \]The first derivative, \( f'(x) \), is calculated using the power rule:\[ f'(x) = \frac{d}{dx} (-x^2) = -2x \]
02

Find the critical point where f'(c) = 0

To find the value of \( c \) where the derivative is zero, set \[ f'(x) = 0 \]So,\[ -2x = 0 \]Solving for \( x \), we get:\[ x = 0 \]Therefore, \( c = 0 \) is the critical point where the first derivative is zero.
03

Determine if f(x) has a local extremum at x=c

To determine if there is a local extremum at \( x = c \), we investigate the second derivative, \( f''(x) \). The function \[ f''(x) = \frac{d}{dx}(-2x) = -2 \]Since \( f''(x) = -2 \) is negative, this indicates that the function is concave down at \( x = 0 \). Hence, \( f(x) \) has a local maximum at \( x = 0 \).

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.

Critical Points
Critical points are essential in calculus, as they help us understand a function's behavior and identify where extreme values occur. A critical point of a function occurs where its derivative, the rate of change of the function, is equal to zero or is undefined. These points are significant because they may correspond to a local extremum or a point where the graph changes direction.

In our exercise, the given function is \[ f(x) = -x^2 \]The first step is to calculate its first derivative:\[ f'(x) = -2x \]The critical point is where this derivative equals zero: \[ -2x = 0 \]Solving for \( x \), we find \( x = 0 \). Hence, \( c = 0 \) is our critical point. This is the point where we need to analyze if there is a local extremum.
Derivative
The derivative of a function provides insight into the rate at which the function's value changes concerning its input. It is a fundamental tool in calculus used for finding slopes of tangent lines, rates of change, and solving optimization problems.

For the given function \[ f(x) = -x^2 \]its first derivative is:\[ f'(x) = -2x \]This derivative tells us how the function \( f(x) \) changes for small changes in \( x \). When the derivative equals zero, it indicates no change at that particular point, suggesting a possible extremum. Calculating derivatives is a crucial step in finding critical points and understanding the behavior of functions.
Local Extremum
A local extremum refers to points in a function where the function reaches a local maximum or minimum value. It is determined by analysis of the critical points and the concavity of the function near these points.

At the critical point \( x = 0 \) of the function\[ f(x) = -x^2 \]we calculated \[ f'(x) = -2x \]and found \( x = 0 \), let's explore the second derivative, \[ f''(x) = -2 \]which remains constant and negative. A negative second derivative indicates that the function is concave down about that point. This curvature indicates a local maximum at \( x = 0 \). Thus, at the critical point \( x = 0 \), the function achieves a local extremum, specifically a local maximum.

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

In Problem 9 we neglected to consider the time delay between a pill being taken and the drug entering the patient's blood. In Chapter 8 we will introduce compartment models as models for drug absorption. We will show that a good model for a drug being absorbed from the gut is that the rate of drug absorption, \(A(t)\), varies with time according to: $$ A(t)=C e^{-k t}, t \geq 0 $$ where \(C>0\) and \(k>0\) are coefficients that will depend on the type of drug, as well as varying between patients. (a) Assume that the drug has first order elimination kinetics, with elimination rate \(k_{1} .\) Show that the amount of drug in the patient's blood will obey a differential equation: $$ \frac{d M}{d t}=C e^{-k t}-k_{1} M $$ (b) Verify that a solution of this differential equation is: $$ M(t)=\frac{C e^{-k t}}{k_{1}-k}+a e^{-k_{1} t} $$ where \(a\) is any coefficient, and we assume \(k_{1} \neq k\). (c) To determine the coefficient \(a\), we need to apply an initial condition. Assume that there was no drug present in the patient's blood when the pill first entered the gut (that is, \(M(0)=0\) ). Find the value of \(a\). (d) Let's assume some specific parameter values. Let \(C=2\), \(k=3\), and \(k_{1}=1 .\) Show that \(M(t)\) is initially increasing, and then starts to decrease. Find the maximum level of drug in the patient's blood. (e) Show that \(M(t) \rightarrow 0\) as \(t \rightarrow \infty\). (f) Using the information from (d) and (e), make a sketch of \(M(t)\) as a function of \(t\).

Find the general antiderivative of the given function. $$ f(x)=\sin (2 x) $$

Find the general antiderivative of the given function. $$ f(x)=\frac{1}{e^{2 x}} $$

Find the general solution of the differential equation. $$ \frac{d y}{d x}=e^{x+1}, x>0 $$

Find the equilibria of $$x_{t+1}=\frac{1}{6}\left(x_{t}^{2}+x_{t}+4\right), \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether they are stable or unstable.
See all solutions

Recommended explanations on Biology Textbooks

Biological Processes

Read Explanation

Responding to Change

Read Explanation

Ecology

Read Explanation

Biological Organisms

Read Explanation

Cells

Read Explanation

Cellular Energetics

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.