/*! 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 46 41-54 Find the absolute maximum ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 46

41-54 Find the absolute maximum and absolute minimum values of f on the given interval. $$f(x)=\left(x^{2}-1\right)^{3}, \quad[-1,2]$$

Short Answer

Expert verified
Max is 27 at x=2; min is -1 at x=0.

Step by step solution

01

Identify the function and interval

The function given is \( f(x) = (x^2 - 1)^3 \), and we need to find the absolute maximum and minimum on the interval \([-1, 2]\).
02

Find the derivative

Differentiate the function to find the critical points. Let \( u = x^2 - 1 \), then \( f(x) = u^3 \). Using the chain rule, the derivative is \( f'(x) = 3u^2 \cdot u' \), where \( u' = 2x \). Thus, \( f'(x) = 3(x^2 - 1)^2 \times 2x = 6x(x^2 - 1)^2 \).
03

Set the derivative to zero

Set \( f'(x) = 6x(x^2 - 1)^2 = 0 \) to find critical points. Solve for \( x \):- \( 6x = 0 \) yields \( x = 0 \)- \( (x^2 - 1)^2 = 0 \) yields \( x^2 - 1 = 0 \), so \( x = \pm 1 \).
04

Evaluate the function at critical points and endpoints

Evaluate \( f(x) \) at \( x = -1, 0, 1, 2 \):- \( f(-1) = ((-1)^2 - 1)^3 = 0^3 = 0 \)- \( f(0) = (0^2 - 1)^3 = (-1)^3 = -1 \)- \( f(1) = (1^2 - 1)^3 = 0^3 = 0 \)- \( f(2) = (2^2 - 1)^3 = 3^3 = 27 \)
05

Determine absolute maximum and minimum

From the evaluations in Step 4, compare values:- The absolute maximum value is \( 27 \) at \( x = 2 \).- The absolute minimum value is \( -1 \) 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
In calculus, critical points occur where the derivative of a function is zero or undefined. These points are essential as they can indicate where a function's behavior changes, such as turning from increasing to decreasing or vice versa. They are important in finding absolute extrema (maximum or minimum values) on a given interval.
A function's critical points are found by first taking the derivative and then setting it equal to zero. In our case, the function is given by \( f(x) = (x^2 - 1)^3 \).
  • First, we find the derivative \( f'(x) \) and set it equal to zero: \( 6x (x^2 - 1)^2 = 0 \).
  • This equation gives us potential points of interest: \( x = 0 \) and \( x = \pm 1 \).
By checking these points further with the original function, we can evaluate any potential maximum or minimum on the given interval.
Derivative
The derivative of a function provides vital information on the rate of change at any given point within the function. It can highlight where a function is increasing or decreasing.
In our exercise, finding the derivative of \( f(x) = (x^2 - 1)^3 \) involves using the chain rule.
  • We first set \( u = x^2 - 1 \), thus transforming the function into \( u^3 \).
  • The derivative \( f'(x) = 3u^2 \cdot (2x) \) simplifies the process.
  • Finally, substituting back, we get \( f'(x) = 6x (x^2 - 1)^2 \).
This derivative helps us establish key points where the slope of the tangent to the curve is zero, indicating critical points.
Interval Evaluation
Interval evaluation is a technique used to examine points across a predefined range to determine the extremal values of a function. It involves calculating the function's value at both the critical points and the interval endpoints. In this case, we evaluate the function \( f(x) = (x^2 - 1)^3 \) at the critical points \( x = -1, 0, 1 \), and the interval endpoints \( x = -1, 2 \):
  • \( f(-1) = ((-1)^2 - 1)^3 = 0 \)
  • \( f(0) = (0^2 - 1)^3 = -1 \)
  • \( f(1) = (1^2 - 1)^3 = 0 \)
  • \( f(2) = (2^2 - 1)^3 = 27 \)
After evaluating, compare these values to identify the absolute maximum \( 27 \) and minimum \( -1 \) on the interval \([-1,2]\).
Chain Rule
The chain rule is a fundamental derivative rule used when differentiating composite functions. A composite function arises when one function is applied to the result of another.In this exercise, our original function \( f(x) = (x^2 - 1)^3 \) involves such a composition. We need to differentiate it using the chain rule.
  • First, identify the inner function and outer function: \( u = x^2 - 1 \), so \( f(u) = u^3 \).
  • Differentiate the outer function \( 3u^2 \), then multiply by the derivative of the inner function \( u' = 2x \), giving \( 6x(x^2 - 1)^2 \).
Applying the chain rule correctly allows us to find the derivative efficiently in problems involving composite functions.

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

$$\begin{array}{l}{\text { Let } f(t) \text { be the temperature at time } t \text { where you live and }} \\ {\text { suppose that at time } t=3 \text { you feel uncomfortably hot. How }} \\ {\text { do you feel about the given data in each case? }}\end{array}$$ $$ \begin{array}{l}{\text { (a) } f^{\prime}(3)=2, \quad f^{\prime \prime}(3)=4} \\\ {\text { (b) } f^{\prime}(3)=2, \quad f^{\prime \prime}(3)=-4} \\ {\text { (c) } f^{\prime}(3)=-2, \quad f^{\prime \prime}(3)=4} \\ {\text { (d) } f^{\prime}(3)=-2, \quad f^{\prime \prime}(3)=-4}\end{array} $$

Solve the initial-value problem. \(\frac{d v}{d t}=e^{-t}\left(1+e^{2 t}\right), \quad t \geqslant 0, \quad v(0)=3\)

Find the equilibria of the difference equation and classify them as stable or unstable. \(x_{t+1}=x_{t}^{3}-3 x_{t}^{2}+3 x_{t}\)

\(37-44\) (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of \(f .\) $$f(x)=e^{\arctan x}$$

Antibiotic pharmacokinetics Suppose that antibiotics are injected into a patient to treat a sinus infection. The antibiotics circulate in the blood, slowly diffusing into the sinus cavity while simultaneously being filtered out of the blood by the liver. In Chapter 10 we will derive a model for the concentration of the antibiotic in the sinus cavity as a function of time since the injection: $$ c(t)=\frac{e^{-\alpha t}-e^{-\beta t}}{\beta-\alpha} \quad \text { where } \beta>\alpha>0 $$ $$ \begin{array}{l}{\text { (a) At what time does } c \text { have its maximum value? }} \\ {\text { (b) At what time does the inflection point occur? What is }} \\ {\text { the significance of the inflection point for the concen- }} \\\ {\text { tration function? }} \\ {\text { (c) Sketch the graph of } c \text { . }}\end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics 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.