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Chapter 4: Problem 103

Use the linear approximation for the function \(f(x)=(1+x)^{n}\) at \(x=0\) and the definition of \(e\) to conclude without using a calculator that \(e>2\).

Short Answer

Expert verified
Since the linear approximation yields 2 at x=1, and e is the limit as n approaches infinity, e > 2.

Step by step solution

01

Recall the linear approximation formula

The linear approximation of a function at a point is given by the formula: \[ f(x) \thickapprox f(a) + f'(a)(x - a) \]In this case, we will use the function \( f(x) = (1 + x)^n \) with \( a = 0 \).
02

Compute the function and its derivative at x=0

First, evaluate the function at \( x = 0 \): \[ f(0) = (1 + 0)^n = 1 \]Next, find the derivative of \( f(x) \): \[ f'(x) = n(1 + x)^{n-1} \]Evaluate the derivative at \( x = 0 \): \[ f'(0) = n(1 + 0)^{n-1} = n \]
03

Formulate the linear approximation

Using the linear approximation formula: \[ f(x) \thickapprox f(0) + f'(0) \times (x - 0) \]Substitute the values found in Step 2: \[ f(x) \thickapprox 1 + n x \]
04

Apply the approximation to conclude about e

For a base value of \(n = 1\), our function simplifies to \( (1 + x)^1 = 1 + x \). Using the definition of \( e \) as the limit when \( x \to 0 \) of \( (1 + x)^{1/x} \), substitute \( x = 1 \): \[ f(1) \thickapprox 1 + 1 = 2 \]Therefore, as \( e \) is the value that \( (1 + 1/n)^n \) approaches as \( n \to \infty \), and since our approximation at \( x = 1 \) yields 2, \( e \) must be greater than 2.

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Key Concepts

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

derivative
The derivative is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. Think of it as the slope of the tangent line to the curve of the function at a specific point. For the function given in the exercise, \( f(x) = (1 + x)^n \), the derivative is found using the power rule. The power rule tells us that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
Using this rule, we differentiate \( f(x) \):\[ f'(x) = n(1 + x)^{n-1} \] The derivative evaluated at \( x = 0 \) is particularly useful for linear approximation:
\[ f'(0) = n(1 + 0)^{n-1} = n \].
function evaluation
Evaluating a function means determining its output for a specific input. In our exercise, we start by evaluating the function \( f(x) = (1 + x)^n \) at \( x = 0 \). Plugging in 0 for x, we have:
\[ f(0) = (1 + 0)^n = 1 \]
This is a simple but crucial step. Understanding this helps us in applying the linear approximation formula more effectively later on. Linear approximation simplifies complex functions and makes calculations easier to handle.
definition of e
The number \( e \) is a special mathematical constant approximately equal to 2.71828. It is the base of the natural logarithm and is critical in calculus and higher mathematics. One way to define \( e \) is through the limit:
\[ e = \lim_{x \to 0} (1 + x)^{1/x} \] This means we consider the base of the exponent increasing infinitesimally small. To understand why \( e > 2 \), we can use approximations. If we let \( x = 1 \), the function \( (1 + x)^{n} \) suggests \( e = (1 + 1) = 2 \). Because \( e \) is a limit as \( n \to \infty \), this approximation further implies \( e > 2 \).
limit
A limit in calculus is the value that a function approaches as the input approaches some value. For example, the limit of \( (1 + x)^{1/x} \) as \( x \to 0 \) is \( e \). Limits help define continuous behavior in functions, and they are essential in calculus. In the exercise, we see the limit concept applied to demonstrate why \( e > 2 \). By analyzing the behavior of the function \( (1 +\frac{1}{n})^n \), it becomes evident that as \( n \) increases, the function approaches \( e \). Using linear approximation for simplicity, with an \( n \) value of 1, we initially get 2, hence establishing that \( e \) must exceed 2.

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