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Chapter 9: Problem 5

$$ f(x)=\ln (1+x) $$

Short Answer

Expert verified
Domain: \( x > -1 \); Increasing and concave down for \( x > -1 \).

Step by step solution

01

Understand the given function

The function provided is \( f(x) = \ln(1+x) \). It is a natural logarithmic function where the logarithm is applied to \( 1 + x \).
02

Domain of the function

Since the logarithmic function \( \ln(y) \) is defined only for \( y > 0 \), we need \( 1+x > 0 \). Thus, the domain of the function is \( x > -1 \).
03

First derivative to find the rate of change

To find the first derivative \( f'(x) \) of \( \ln(1+x) \), we apply the chain rule. Since the derivative of \( \ln(u) \) is \( \frac{1}{u} \), we have:\[ f'(x) = \frac{1}{1+x} \cdot 1 = \frac{1}{1+x} \]
04

Critical points and intervals of increase/decrease

The derivative tells us the rate of change of the function. Since \( \frac{1}{1+x} \) is positive for \( x > -1 \), the function is increasing in its entire domain \( x > -1 \).
05

Second derivative to analyze concavity

Calculate the second derivative \( f''(x) \) to understand concavity:\[ f''(x) = \frac{d}{dx} (\frac{1}{1+x}) = -\frac{1}{(1+x)^2} \]This results in a negative value for all \( x > -1 \), indicating the function is concave downward across its domain.
06

Sketch and behavior analysis

The function \( \ln(1+x) \) is defined for \( x > -1 \), increasing over its entire domain, and concave down. It approaches infinity as \( x \) approaches infinity and has a vertical asymptote at \( x = -1 \).

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Key Concepts

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

Domain of a Function
The domain of a function is the set of all possible input values (usually represented by 'x') that will result in a valid output for the function. For natural logarithmic functions, the function is only defined when the argument (the part inside the logarithm) is greater than zero. In our given function, \[ f(x) = \ln(1+x) \]the argument is \[ 1 + x \]. To find the domain, we solve the inequality: \[ 1 + x > 0 \]. Thus, \[ x > -1 \]. This means the domain includes all x-values greater than -1, excluding x = -1 itself since it would make the argument equal to zero, which is undefined in logarithms.
  • Remember: Logarithmic functions require a positive argument.
  • Check the argument to determine the domain.
Identifying the domain is crucial as it tells us the range within which the function can safely be evaluated.
Derivative
The derivative of a function gives us important information about the rate at which the function's values change with respect to changes in its input. For the natural logarithmic function \( f(x) = \ln(1+x) \), the first derivative \( f'(x) \) tells us how the function is increasing or decreasing.
In this case, the derivative is calculated using the chain rule, and for our function, it simplifies to: \[ f'(x) = \frac{1}{1+x} \]. This derivative is positive when \( x > -1 \) because the denominator \( 1+x \) is positive, ensuring that \( f(x) \) is increasing for all x in its domain:
  • The function is always increasing as x increases.
Understanding the derivative is essential, especially when analyzing the behavior of the function such as identifying intervals of increase or decrease and locating critical points.
Concavity
Concavity describes the direction in which a function curves. To determine this for \( f(x) = \ln(1+x) \), we calculate the second derivative.
Computing the second derivative, we find: \[ f''(x) = -\frac{1}{(1+x)^2} \]. Since this expression is negative for all \( x > -1 \), it tells us that the function is concave downward in its entire domain:
  • Concave downward: The curve bends downwards.
Visually, a concave downward shape appears like an upside-down bowl across the domain. This behavior can give insights on the potential behavior of functions, for example, indicating that its rate of increase is slowing down as x gets larger.

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Most popular questions from this chapter

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(1, \frac{2}{2^{2}-1^{2}}, \frac{3}{3^{2}-2^{2}}, \frac{4}{4^{2}-3^{2}}, \ldots\)

Let $$ f(x)= \begin{cases}e^{-1 / x^{2}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases} $$ (a) Show that \(f^{\prime}(0)=0\) by using the definition of the derivative. (b) Show that \(f^{\prime \prime}(0)=0\). (c) Assuming the known fact that \(f^{(n)}(0)=0\) for all \(n\), find the Maclaurin series for \(f(x)\). (d) Does the Maclaurin series represent \(f(x)\) ? (e) When \(a=0\), the formula in Theorem \(\mathrm{B}\) is called Maclaurin's Formula. What is the remainder in Maclaurin's Formula for \(f(x)\) ? This shows that a Maclaurin series may exist and yet not represent the given function (the remainder does not tend to 0 as \(n \rightarrow \infty)\).

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{n \cos (n \pi)}{2 n-1}\)

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1-\frac{x}{1 \cdot 3}+\frac{x^{2}}{2 \cdot 4}-\frac{x^{3}}{3 \cdot 5}+\frac{x^{4}}{4 \cdot 6}-\cdots $$

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{\ln n}{\sqrt{n}}\)
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