/*! 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 25 Find the exact location of all t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 25

Find the exact location of all the relative and absolute extrema of each function. \(h(x)=(x-1)^{2 / 3}\) with domain [0,2]

Short Answer

Expert verified
The absolute extrema of the function \(h(x)=(x-1)^{2/3}\) with domain [0,2] are at the endpoints of the domain, with \(h(0)=h(2)=1\). There are no relative extrema since there are no critical points.

Step by step solution

01

Find the first derivative of the function

To find the first derivative of the function, apply the power rule as follows: \[h'(x) = \frac{d}{dx} \left( (x-1)^{2/3} \right) = \frac{2}{3}(x-1)^{-1/3}\]
02

Find the critical points of the function

To find the critical points of the function, set the first derivative equal to zero and solve for x: \[h'(x)=\frac{2}{3}(x-1)^{-1/3}=0\] Since the first derivative is never zero for this function, it has no critical points. However, we need to consider the endpoints of the domain as potential extrema.
03

Check the second derivative at the critical points

Find the second derivative of the function: \[h''(x) = \frac{d^2}{dx^2} \left( (x-1)^{2/3} \right) = \frac{2}{9}(x-1)^{-4/3}\] Since there are no critical points, we don't need to evaluate the second derivative at any points.
04

Calculate the function value at the endpoints

To find the absolute extrema of the function, we need to evaluate the function at the endpoints of the given domain: \[h(0) = (0-1)^{2/3} = 1\] \[h(2) = (2-1)^{2/3} = 1\] There are no relative extrema in this function since there are no critical points. From our analysis, we can conclude that the absolute extrema are at the endpoints of the domain, with \(h(0) = h(2) = 1\).

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.

First Derivative Test
When you need to identify potential extrema (maxima or minima) in calculus, the First Derivative Test is an essential tool. The first derivative of a function, denoted as \( f'(x) \), provides us with information about the slope of the function. By applying it, we can check where the slope changes from positive to negative or vice versa.
The First Derivative Test involves:
  • Finding the first derivative \( h'(x) \) of the function, by applying rules like the power rule.
  • Setting \( h'(x) = 0 \) to find critical points where the slope is zero or undefined, which might indicate a local extremum.
  • Analyzing the sign changes around these points to ascertain if they are indeed maxima or minima.
However, if the first derivative doesn't become zero, as seen in our function \( h(x)=(x-1)^{2/3} \), we check endpoints for extrema. The derivative \( h'(x) \) in this case pointed out no critical points within the domain [0,2].
Critical Points
Critical points are specific points in the domain of a function where the first derivative is zero or undefined. These points are significant because they can indicate potential locations of relative maxima or minima.
Here鈥檚 how you typically find them:
  • Compute the first derivative \( h'(x) \).
  • Solve \( h'(x) = 0 \) to find any points where the slope equals zero.
  • Determine if \( h'(x) \) is undefined; this could also provide critical points.
In the case of our function, \( h(x)=(x-1)^{2/3} \), the derivative does not become zero, and it is undefined at \( x = 1 \). This points to checking the domain's endpoints, as the original function is continuous everywhere else within the interval.
Second Derivative Test
The Second Derivative Test helps in determining the concavity of a function at critical points, which in turn reveals if the point is a local maxima or minima. After identifying critical points using the first derivative:
  • Find the second derivative, \( h''(x) \).
  • Evaluate the second derivative at each critical point.
  • Check the sign of \( h''(x) \):
    • If \( h''(x) > 0 \), the function is concave up, indicating a local minimum.
    • If \( h''(x) < 0 \), the function is concave down, implying a local maximum.
    • If \( h''(x) = 0 \), the test is inconclusive, and further analysis is needed.
In our example, since there were no critical points from our first derivative, we did not need the second derivative to determine extremum locations.
Absolute Extrema
Absolute extrema refer to the highest or lowest points on a function over a specific interval. Finding them involves checking both critical points within the domain and the endpoints of the interval itself.
Evaluating absolute extrema requires:
  • Determining all critical points where \( h'(x)=0 \) or undefined.
  • Calculating the function values at these critical points.
  • Checking function values at the interval's endpoints.
For the function \( h(x)=(x-1)^{2/3} \) over [0, 2], since no critical points exist, the extrema were checked at the endpoints. As calculated, \( h(0) = 1 \) and \( h(2) = 1 \), indicating that these points are the absolute extrema. The function does not have relative extrema, as the slope doesn't change signs within the open 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

Refer back to the model in the preceding exercise. Assume that someone has completed 14 years of school and that her income is increasing by \(\$ 10,000\) per year. How much schooling per year is this rate of increase equivalent to?

Your friend tells you that he has found a continuous function defined on \((-\infty,+\infty)\) with exactly two critical points, each of which is a relative maximum. Can he be right?

The weekly sales of Honolulu Red Oranges is given by \(q=1,000-20 p\). Calculate the price elasticity of demand when the price is \(\$ 30\) per orange (yes, \(\$ 30\) per orange \(^{63}\) ). Interpret your answer. Also, calculate the price that gives a maximum weekly revenue, and find this maximum revenue. HINT [See Example 1.]

You have been hired as a marketing consultant to Big Book Publishing, Inc., and you have been approached to determine the best selling price for the hit calculus text by Whiner and Istanbul entitled Fun with Derivatives. You decide to make life easy and assume that the demand equation for Fun with Derivatives has the linear form \(q=m p+b\), where p is the price per book, q is the demand in annual sales, and m and b are certain constants you'll have to figure out. a. Your market studies reveal the following sales figures: when the price is set at $$\$ 50.00$$ per book, the sales amount to 10,000 per year; when the price is set at $$\$ 80.00$$ per book, the sales drop to 1000 per year. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes annual revenue and predict what Big Book Publishing, Inc.'s annual revenue will be at that price.

By thinking about extrema, show that, if \(f(x)\) is a polynomial, then between every pair of zeros \((x\) -intercepts) of \(f(x)\) there is a zero of \(f^{\prime}(x)\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Pure 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.