/*! 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 27 Find the exact location of all t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
In conclusion, the function \(k(x)=\frac{2x}{3}+(x+1)^{2/3}\) has an absolute maximum located at \(x = 0\) with a function value of \(1\). There are no other relative or absolute extrema within the given domain \((-\infty, 0]\).

Step by step solution

01

Calculate the derivative of the function

To find the critical points of the function, we first need to calculate its derivative. We can do this using the sum/difference rule and the chain rule. The derivative of \(k(x)=\frac{2x}{3}+(x+1)^{2/3}\) is: \[k'(x) = \frac{d}{dx} \left(\frac{2x}{3}\right) + \frac{d}{dx} \left((x+1)^{2/3}\right)\] \[k'(x) = \frac{2}{3} + \frac{2}{3}(x+1)^{-1/3}\]
02

Find the critical points

Now that we have the derivative, we can find the critical points by setting it equal to 0 and finding for which values of \(x\) the derivative is undefined. Equating the derivative to 0, we get: \[\frac{2}{3} + \frac{2}{3}(x+1)^{-1/3} = 0\] To solve this equation, we can start by isolating the term involving \(x\): \[(x+1)^{-1/3} = -1\] Next, we can invert both sides (raising them to the power of -1): \[(x+1)^{1/3} = 1\] Now, we can simply cube both sides to isolate \(x\): \[x+1 = 1^3\] \[x = 0\] Thus, there is one critical point at \(x = 0\). We also need to check whether the derivative is undefined. Since the derivative is a continuous function, we can conclude that there are no values of \(x\) for which the derivative is undefined within the given domain.
03

Test the critical points and endpoints

Now that we have the critical point \(x=0\), we need to determine if it is a relative extrema. It may be useful to create a sign chart for \(k'(x)\) to see how the function behaves around this critical point: 1. \(k'(x) > 0\) for \(x < 0\). 2. \(k'(x) < 0\) for \(x > 0\). Since the domain of the function only includes the interval \((-\infty, 0]\), the fact that the derivative changes from positive to negative at the critical point \(x = 0\) indicates that there is a relative maximum at this critical point. The function value at this relative maximum is: \[k(0) = \frac{2(0)}{3} + (0 + 1)^{2/3} = 1\] Since the relative maximum is the only extremum within the domain, it is also the absolute maximum. Now, we need to check the endpoint \(x = 0\). We already know the value of the function at \(x = 0\): \(k(0) = 1\). No further analysis of the endpoint is needed as no other potential extremum is present within the domain.
04

Summarize the findings

In conclusion, the function \(k(x)=\frac{2x}{3}+(x+1)^{2/3}\) has an absolute maximum located at \(x = 0\) with a function value of \(1\). There are no other relative or absolute extrema within the given domain.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
In the realm of calculus, critical points are where a function's graph can have key changes in direction, potentially indicating a relative maximum or minimum. These points occur where the function's derivative is either equal to zero or undefined. To find critical points, we first need the derivative of the function, which gives us the rate at which the function is changing at any given point.

When dealing with a function like \(k(x)=\frac{2x}{3}+(x+1)^{2/3}\), we identify critical points by setting the derivative \(k'(x)\) equal to zero and solving for \(x\), or by finding values where the derivative does not exist. In our example, through some algebraic manipulation, the critical point is found at \(x=0\), which happens to be within the function’s domain, \((-fty, 0]\). While looking for such points, it's important to cross-reference them with the domain to ensure they're actually relevant for our function. A critical point outside of the domain would not impact the function's graph.
Derivative of a Function
The derivative serves as a cornerstone concept in calculus. It's a mathematical tool that provides, at any given point, the rate of change or the slope of the graph of a function. When we say we're taking the derivative, or calculating \(k'(x)\) for our function \(k(x)\), we're essentially finding a new function that quantifies how \(k(x)\) speeds up or slows down, and in which direction it moves.

In the context of the function \(k(x)=\frac{2x}{3}+(x+1)^{2/3}\), we use rules such as the sum rule, the power rule, and the chain rule to differentiate each term separately. The sum rule allows us to split the derivative into parts, the power rule simplifies the differentiation of powers of \(x\), and the chain rule handles compositions of functions, as seen with \((x+1)^{2/3}\). Determining the derivative correctly is crucial. A single mistake can lead us to incorrect critical points, thereby misguiding our analysis of the function's extrema.
Relative Maximum
The peak of a hill on the graph of a function represents a relative maximum. It's a point where the function value is higher than all the nearby points. For functions defined over an interval, a relative maximum is precisely where the function transitions from increasing to decreasing, implying that the derivative goes from positive to negative. Establishing the existence of a relative maximum (or minimum) can imply key behaviors about the function over a specific interval.

For our example function \(k(x)\), the critical point \(x = 0\) is tested to determine if it's a relative maximum. By understanding the behavior of \(k'(x)\) around \(x = 0\)—positive before and negative after—we conclude that there is indeed a relative maximum at this point. Since the domain restricts us to \((-fty, 0]\), and given that there are no other critical points nor discontinuities, this relative maximum is also the absolute maximum of the function within the domain.

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

Daily sales of Kent's Tents reached a maximum in January 2002 and declined to a minimum in January 2003 before starting to climb again. The graph of daily sales shows a point of inflection at June 2002 . What is the significance of the point of inflection?

You manage a small antique store that owns a collection of Louis XVI jewelry boxes. Their value \(v\) is increasing according to the formula $$v=\frac{10,000}{1+500 e^{-0.5 t}}$$ where \(t\) is the number of years from now. You anticipate an inflation rate of \(5 \%\) per year, so that the present value of an item that will be worth v in t years' time is given by $$p=v(1.05)^{-t}$$ What is the greatest rate of increase of the value of your antiques, and when is this rate attained?

The percentage of U.S.-issued mortgages that were subprime can be approximated by $$A(t)=\frac{15.0}{1+8.6(1.8)^{-t}} \text { percent } \quad(0 \leq t \leq 8)$$ \(t\) years after the start of \(2000^{52}\) Graph the derivative \(A^{\prime}(t)\) of \(A(t)\) using an extended domain of \(0 \leq t \leq 15\). Determine the approximate coordinates of the maximum and determine the behavior of \(A^{\prime}(t)\) at infinity. What do the answers tell you?

A general linear demand function has the form \(q=m p+b(m\) and \(b\) constants, \(m \neq 0\) ). a. Obtain a formula for the price elasticity of demand at a unit price of \(p\). b. Obtain a formula for the price that maximizes revenue.

Your calculus study group is discussing price elasticity of demand, and a member of the group asks the following question: "Since elasticity of demand measures the response of demand to change in unit price, what is the difference between elasticity of demand and the quantity \(-d q / d p ? "\) How would you respond?

See all solutions

Recommended explanations on Math Textbooks

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.