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Chapter 5: Problem 4

Find the local maxima and minima of each of the functions. Determine whether each function has local maxima and minima and find their coordinates. For each function, find the intervals on which it is increasing and the intervals on which it is decreasing. \(y=\ln \left(\frac{x}{x+1}\right), x>0\)

Short Answer

Expert verified
No local maxima or minima, function is increasing on \((0, \infty)\).

Step by step solution

01

Find the derivative

To find local maxima and minima, we first need the derivative of the function. For the function \( y = \ln \left( \frac{x}{x+1} \right) \), use the quotient rule and the chain rule to differentiate:\[\frac{d}{dx} [y] = \frac{1}{\frac{x}{x+1}} \cdot \frac{d}{dx} \left(\frac{x}{x+1}\right)\]This simplifies to:\[\frac{d}{dx} \left( \frac{x}{x+1} \right) = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2}\]So, the derivative is:\[\frac{d}{dx} [y] = \frac{1}{\frac{x}{x+1}} \cdot \frac{1}{(x+1)^2} = \frac{x+1}{x} \cdot \frac{1}{(x+1)^2} = \frac{1}{x(x+1)}\]
02

Find critical points

To find critical points, set the derivative equal to zero and solve for \(x\):\[\frac{1}{x(x+1)} = 0\]Since the numerator is \(1\) and not zero, the derivative doesn't equal zero for any value of \(x > 0\). However, critical points can also occur where the derivative is undefined. The derivative is undefined at \(x = 0\), but \(x > 0\), so there are no critical points within the domain.
03

Determine intervals of increasing/decreasing

Since there are no critical points where the derivative is zero, examine the sign of the derivative:\[\frac{1}{x(x+1)} > 0 \text{ for } x > 0\]The derivative is positive for all \(x > 0\), thus the function is increasing on the interval \((0, \infty)\). There are no intervals where the function is decreasing.
04

Identify local maxima and minima

We have established that the function is always increasing for \(x > 0\). A function that is always increasing does not have any local maxima or minima. Therefore, there are no local extrema in the given domain.

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Key Concepts

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

Derivative
The derivative of a function gives us vital information about the rate of change of the function's output concerning changes in the input. It tells how steeply the function is climbing or descending.

In this problem, we started by finding the derivative of the function \( y = \ln \left( \frac{x}{x+1} \right) \). To do so, we used both the quotient rule and the chain rule. These powerful techniques help when you have complex functions that involve one function inside another or fractions of functions.

The result of applying these rules gave us the derivative \( \frac{1}{x(x+1)} \), which is crucial for the following steps. This derivative tells us how the function behaves over its entire domain, \( x > 0 \). By understanding the nature of derivatives, we gain insights into the overall behavior of the function.
Local Maxima and Minima
Local maxima and minima refer to the highest or lowest points, respectively, within a particular interval. They represent critical spots on the graph where a switch from increasing to decreasing occurs or vice versa.

For our given function, we determined there are no local maxima or minima. This is because the function is consistently increasing across its domain and does not have any unique peaks or valleys.

A critical characteristic to note is that when a function doesn't change direction, as in this case, it typically won't have any local extremes. Thus, always checking both the derivative and the interval helps ensure that any extremum is not missed.
Critical Points
Critical points are typically where the derivative equals zero or is undefined. These points often indicate where a function might reach a local maximum or minimum.

In our function's case, we set the derivative, \( \frac{1}{x(x+1)} \), to zero. Finding no solution within our domain \( x > 0 \) implies no critical points are arising from where the derivative equals zero.

The derivative is undefined at \( x = 0 \), but since this point lies outside the domain, it does not affect the function's critical points within the relevant interval. Understanding critical points is essential as they are the checkpoints that highlight potential changes in the function's direction.
Increasing and Decreasing Intervals
Increasing and decreasing intervals give insights into the overall behavior of a function. If a function's derivative is positive over an interval, the function is increasing there. Conversely, if the derivative is negative, the function is decreasing.

For our function \( y = \ln \left( \frac{x}{x+1} \right) \), we found the derivative, \( \frac{1}{x(x+1)} \), to be positive for all \( x > 0 \), indicating that the function is increasing over its entire domain. There are no intervals where the function is decreasing.

Detecting these intervals is crucial as they help describe the flow and changes in the function, highlighting areas of growth or decline and supporting our understanding of the function's overall behavior.

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Most popular questions from this chapter

The growth of a particular population is described by a power law model, in which the rate of growth is given by a function: $$ r(t)=\frac{A}{(t+a)^{m}} $$ where \(A, m\), and \(a\) are all unknown constants. Given the following data for the size of the population, calculate the value for these constants that would fit the model to the data: $$ \begin{array}{ll} \hline \boldsymbol{t} & \boldsymbol{r}(\boldsymbol{t}) \\ \hline 0 & 1.89 \\ 1 & 1.31 \\ 3 & 0.988 \\ \hline \end{array} $$ Hint: Eliminate \(A\) first. It may help to then take logarithms of the equations that you derive after eliminating \(A\).

Find the general antiderivative of the given function. $$ f(x)=\frac{3}{e^{-x}} $$

Solve the initial-value problem. $$ \frac{d y}{d x}=\sqrt{x}, \text { for } x \geq 0 \text { with } y(1)=2 $$

Initially you measure that a colony of bacterial cells contains 2000 cells. 2 hours later you measure the colony again, and count 4000 cells. (a) How many cells would you expect the colony to contain 3 hours after the start of the experiment? (b) In fact, you realize that the hemocytometer that you used to count the cells for both measurements is only accurate to \(10 \%\), meaning that if you count 1000 cells, the real number of cells is somewhere between \(1000-100=900\) cells and \(1000+100=\) 1100 cells. What is the largest possible number of cells in the colony 3 hours after the start of the experiment? And what is the smallest possible number of cells at 3 hours?

The dynamics of a population of fish is modeled using the Beverton-Holt model: $$N_{t+1}=\frac{3 N_{t}}{1+\frac{N_{t}}{30}}$$ (a) Calculate the first ten terms of the sequence when \(N_{0}=10\). (b) Calculate the first ten terms of the sequence when \(N_{0}=120\). (c) Find all equilibria of the system, and use the stability criterion to determine which of them (if any) are stable. (d) Explain why your answers from (a) and (b) are consistent with what you have determined about the equilibria of the system.
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