/*! 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 Identify the critical points. Th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum. $$ f(x)=\frac{\ln (x+2)}{x+2} $$

Short Answer

Expert verified
The function has a local maximum at \( x = e-2 \).

Step by step solution

01

Find the First Derivative

First, we need to find the derivative of the function \( f(x) = \frac{\ln(x+2)}{x+2} \). We use the quotient rule \( \frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{u'v - uv'}{v^2} \). Let \( u = \ln(x+2) \) and \( v = x+2 \):\( u' = \frac{1}{x+2} \, (u = \ln(x+2)), \quad v' = 1 \, (v = x+2) \)Apply the quotient rule:\( f'(x) = \frac{\frac{1}{x+2} \cdot (x+2) - \ln(x+2) \cdot 1}{(x+2)^2} = \frac{1 - \ln(x+2)}{(x+2)^2}\)
02

Find the Critical Points

A critical point occurs where the derivative \( f'(x) \) is zero or undefined. Since we found that:\( f'(x) = \frac{1 - \ln(x+2)}{(x+2)^2}\)Set the numerator equal to zero:\( 1 - \ln(x+2) = 0 \)Solving for \(x\):\( \ln(x+2) = 1 \)Exponentiate both sides:\( x+2 = e^1 \)\( x = e - 2 \)The critical points are where \( x = e - 2 \). The denominator \((x+2)^2\) is always positive except where \(x = -2\), hence no critical points from undefined derivative aside \(x > -2\).
03

Apply the First Derivative Test

To apply the First Derivative Test, examine the sign of \( f'(x) \) to the left and right of the critical point \( x = e - 2 \).For \( x < e - 2 \), choose \( x = 0 \), then \( \ln(x+2) = \ln(2) < 1 \), so \( f'(x) > 0 \).For \( x > e - 2 \), choose \( x = 1 \), then \( \ln(x+2) = \ln(3) > 1 \), so \( f'(x) < 0 \).This change from positive to negative indicates a local maximum at \( x = e - 2 \).
04

Apply the Second Derivative Test

To further verify, we use the Second Derivative Test. First, find the second derivative \( f''(x) \) by differentiating \( f'(x) = \frac{1 - \ln(x+2)}{(x+2)^2} \).Differentiate using the quotient rule again:Let \( g = 1 - \ln(x+2) \), \( h = (x+2)^2 \)\( g' = -\frac{1}{x+2}, \quad h' = 2(x+2)\)The second derivative is:\( f''(x) = \frac{(-1/(x+2))\cdot (x+2)^2 - (1 - \ln(x+2))\cdot 2(x+2)}{(x+2)^4}\)Simplifying and substituting \( x = e-2 \), \( f''(x) = - \frac{(e-2+2)^2 + 2(e-1)(e-2+2)}{(e)^4} < 0 \)This confirms a local maximum at \( x = e-2 \) as \( f''(x) < 0 \). Therefore, the point \( x = e-2 \) is a local maximum.

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.

First Derivative Test
The First Derivative Test is a handy tool used in calculus to determine whether a critical point is a local maximum or minimum. A critical point is found where the first derivative of a function, denoted as \( f'(x) \), is zero or undefined. Here is how the test works:
  • Arrange intervals around the critical point and
  • examine the sign of \( f'(x) \) on each interval.
If the derivative changes from positive to negative as it passes through the critical point, then you have a local maximum at that point. Conversely, if it changes from negative to positive, the point is a local minimum. For the function given by \( f(x) = \frac{\ln(x+2)}{x+2} \), the critical point was found to be \( x = e - 2 \). By choosing values less than \( e - 2 \) and greater than \( e - 2 \), you can determine the behavior of \( f'(x) \):
  • For \( x < e - 2 \), \( f'(x) > 0 \), indicating increasing behavior.
  • For \( x > e - 2 \), \( f'(x) < 0 \), showing decreasing behavior.
Thus, the function increases to the left of \( e - 2 \) and decreases to the right, indicating that \( x = e - 2 \) is a local maximum.
Second Derivative Test
The Second Derivative Test provides another way to determine whether a critical point is a local maximum or minimum. This test examines the second derivative of the function, \( f''(x) \), at the critical point. The steps to apply this test are simple:
  • Find \( f''(x) \), the second derivative of the function.
  • Evaluate \( f''(x) \) at the critical point.
If \( f''(x) > 0 \) at the critical point, the point is a local minimum as the graph shows concave up behavior. If \( f''(x) < 0 \), then it is a local maximum due to the concave down behavior of the function. In our example of \( f(x) = \frac{\ln(x+2)}{x+2} \), the second derivative evaluated at \( x = e - 2 \) is negative \( f''(e-2) < 0 \).This means that the function graph is curving downwards at this point, confirming the local maximum we found using the First Derivative Test.
Quotient Rule in Calculus
The Quotient Rule is essential in calculus for finding the derivative of a function that is the ratio of two differentiable functions. This rule states:\[\frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{u'v - uv'}{v^2}\]Here, \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives. The rule ensures that you differentiate the numerator and denominator separately before combining them using the formula. In practice, you:
  • Calculate the derivative of the numerator \( u' \) while keeping the denominator unchanged.
  • Subtract the product of the original numerator \( u \) and the derivative of the denominator \( v' \).
  • Divide the entire expression by the square of the denominator \( v^2 \).
Applying the Quotient Rule to the function \( f(x) = \frac{\ln(x+2)}{x+2} \) allowed us to find the first derivative, \( f'(x) = \frac{1 - \ln(x+2)}{(x+2)^2} \), critical for locating critical points and analyzing the function's behavior.

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

Suppose that every customer order taken by the XYZ Company requires exacty 5 hours of labor for handling the paperwork; this length of time is fixed and does not vary from lot to lot. The total number of hours \(y\) required to manufacture and sell a lot of size \(x\) would then be \(y=(\) number of hours to produce a lot of size \(x)+5\) Some data on XYZ's bookcases are given in the following table. $$ \begin{array}{ccc} \hline & & \text { Total Labor } \\ \text { Order } & \text { Lot Size } x & \text { Hours } y \\ \hline 1 & 11 & 38 \\ 2 & 16 & 52 \\ 3 & 8 & 29 \\ 4 & 7 & 25 \\ 5 & 10 & 38 \\ \hline \end{array} $$ (a) From the description of the problem, the least-squares line should have 5 as its \(y\) -intercept. Find a formula for the value of the slope \(b\) that minimizes the sum of squares $$ S=\sum_{i=1}^{n}\left[y_{i}-\left(5+b x_{i}\right)\right]^{2} $$ (b) Use this formula to estimate the slope \(b\). (c) Use your least-squares line to predict the total number of labor hours to produce a lot consisting of 15 bookcases.

For the price function defined by $$ p(x)=(182-x / 36)^{1 / 2} $$ find the number of units \(x_{1}\) that makes the total revenue a maximum and state the maximum possible revenue. What is the marginal revenue when the optimum number of units, \(x_{1}\), is sold?

A rectangle has two corners on the \(x\) -axis and the other two on the curve \(y=\cos x\), with \(-\pi / 2

Assume that (1) world population continues to grow exponentially with growth constant \(k=0.0132,(2)\) it takes \(\frac{1}{2}\) acre of land to supply food for one person, and (3) there are \(13,500,000\) square miles of arable land in the world. How long will it be before the world reaches the maximum population? Note: There were \(6.4\) billion people in 2004 and 1 square mile is 640 acres.

If the brakes of a car, when fully applied, produce a constant deceleration of 11 feet per second per second, what is the shortest distance in which the car can be braked to a halt from a speed of 60 miles per hour?

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.