/*! 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 10 Find the absolute maximum and mi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 10

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=\left(x^{2}+x\right)^{2 / 3} ;[-2,3]$$

Short Answer

Expert verified
The absolute minimum is 0 (at \( x = -1, 0, -\frac{1}{2} \)), and the maximum is \( 12^{2/3} \) (at \( x = 3 \)).

Step by step solution

01

Define the problem and identify endpoints

We need to find the absolute maximum and minimum of the function \( f(x) = (x^2 + x)^{2/3} \) on the closed interval \([-2, 3]\). Since the interval is closed, both endpoints \( x = -2 \) and \( x = 3 \) must be considered.
02

Find the derivative

To find critical points, we'll differentiate the function. Use the chain rule to differentiate \( f(x) = (x^2 + x)^{2/3} \). We have:\[ f'(x) = \frac{2}{3}(x^2 + x)^{-1/3} \cdot (2x + 1). \]
03

Set the derivative equal to zero

We find the critical points by setting \( f'(x) = 0 \):\[ \frac{2}{3}(x^2 + x)^{-1/3} \cdot (2x + 1) = 0. \]This equation is zero when \( 2x + 1 = 0 \), which simplifies to \( x = -\frac{1}{2} \).
04

Determine where f'(x) does not exist

The derivative \( f'(x) \) is undefined where \( (x^2 + x)^{-1/3} \) is undefined. Since this occurs when \( x^2 + x = 0 \), solve:\[ x(x + 1) = 0 \] gives \( x = 0 \) or \( x = -1 \).
05

Evaluate the function at critical values and endpoints

Evaluate \( f(x) \) at the critical points and endpoints:- \( f(-2) = ((-2)^2 - 2)^{2/3} = (2)^{2/3} = 2^{2/3} \).- \( f(-1) = 0^{2/3} = 0 \).- \( f(0) = (0^2 + 0)^{2/3} = 0 \).- \( f\left(-\frac{1}{2}\right) = \left(\left(-\frac{1}{2}\right)^2 - \frac{1}{2}\right)^{2/3} = (0)^{2/3} = 0 \).- \( f(3) = ((3)^2 + 3)^{2/3} = (12)^{2/3} \).
06

Compare function values to find maximum and minimum

Compare all evaluated values:- The minimum value is 0, occurring at \( x = -1, 0, -\frac{1}{2} \).- The maximum value is \( 12^{2/3} \), occurring at \( x = 3 \).

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.

Absolute Extrema
Absolute extrema refer to the highest and lowest values that a function attains over a specific interval. This involves identifying the absolute maximum and minimum.
  • The absolute maximum is the largest value of the function within the interval.
  • The absolute minimum is the smallest value of the function within the interval.
To find these extrema, it's crucial to evaluate both the endpoints and any critical points of the function. On a closed interval like o]\[-2, 3\], the process ensures that the global extrema are identified with certainty.
Critical Points
Critical points occur where the derivative of a function is zero or undefined. These are points where the function can potentially have a local maximum, minimum, or a point of inflection. For the function \(f(x) = (x^2 + x)^{2/3}\), we found critical points by setting the derivative equal to zero and analyzing where the derivative is undefined. The equation \(f'(x) = \frac{2}{3}(x^2 + x)^{-1/3} \cdot (2x + 1)\)leads to critical points at \(x = -\frac{1}{2}, 0, \text{ and } -1\),rate the critical points early in processes like finding extrema.
Chain Rule
The chain rule is a fundamental tool in calculus for differentiating compositions of functions. It's especially helpful when dealing with expressions raised to a power, like \(f(x) = (x^2 + x)^{2/3}\).To apply the chain rule, you identify two functions: an outer function and an inner function.
  • In our example, the outer function is \(g(u) = u^{2/3}\), and the inner function is \(u = x^2 + x\).
  • The derivative then involves differentiating both the outer function with respect to the inner function and the inner function with respect to \(x\).
Hence, the derivative \(f'(x) = \frac{2}{3}(x^2 + x)^{-1/3} \cdot (2x + 1)\) uses the chain rule correctly.
Interval Evaluation
Interval evaluation involves assessing a function's behavior at specific points within a defined domain, such as endpoints and critical points. When solving for absolute extrema, evaluating the function at critical points and the endpoints of the interval is key.
  • This process ensures that all potential sites for relative and absolute extrema are considered.
  • Using the interval \([-2, 3]\), you calculate the function \(f(x)\) at \(x = -2, -1, 0, -\frac{1}{2},\) and \(3\) to find the minimum and maximum values.
After evaluating these points, you can confidently determine where the highest and lowest values of a function occur within a given interval.

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

Use a graphing utility to determine the number of times the curves intersect: and then apply Newton's Method, where necded, to approximate the \(x\) -coordinates of all intersections. $$y=x^{2} \text { and } y=\sqrt{2 x+1}$$

Suppose that the equations of motion of a paper airplane during the first 12 seconds of flight are $$x=t-2 \sin t, \quad y=2-2 \cos t \quad(0 \leq t \leq 12)$$ What are the highest and lowest points in the trajectory, and when is the airplane at those points?

Use a CAS to show that the absolute minimum value of $$f(x)=x^{2}+\frac{16 x^{2}}{(8-x)^{2}}, \quad x>8$$ occurs at \(x=4(2+\sqrt[3]{2})\) by using it to find \(f^{\prime}(x)\) and to solve the equation \(f^{\prime}(x)=0\)

Find the absolute maximum and minimum values of \(f\), if any, on the given interval, and state where those values occur. $$f(x)=x^{4}+4 x ;(-\infty,+\infty)$$

Use a graphing utility to determine the number of times the curves intersect: and then apply Newton's Method, where necded, to approximate the \(x\) -coordinates of all intersections. $$y=x^{3} \text { and } y=\frac{1}{3} x-1$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

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.