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Graph \(f\) on the interval \((0,200] .\) Find an approximate equation for the horizontal asymptote. $$f(x)=\left(1+\frac{1}{x}\right)^{x}$$

Short Answer

Expert verified
The horizontal asymptote is approximately \(y = e\) or \(y = 2.72\).

Step by step solution

01

Understanding the Function

The function given is \(f(x) = \left(1 + \frac{1}{x}\right)^{x}\). This function resembles a key form of the exponential function known as Euler's number when \(x\) approaches infinity.
02

Identify Function Behavior as \(x\) Approaches Infinity

As \(x\) becomes very large, the term \(\frac{1}{x}\) becomes very small. Consequently, \(f(x)\) approaches \(e\), where \(e \approx 2.71828\). This is because \(\left(1 + \frac{1}{x}\right)^{x}\) approximates \(e\) when \(x \to \infty\).
03

Graphing on the Interval \((0,200]\)

Plot \(f(x)\) for \(x\) values in the interval \((0, 200]\). As \(x\) increases from 1 to 200, the function \(\left(1 + \frac{1}{x}\right)^{x}\) smoothly increases and converges towards \(e\), becoming very close to \(e\) when \(x\) is high.
04

Determine the Horizontal Asymptote

Considering the behavior as \(x\) approaches 200, for large values of \(x\), \(f(x)\) is very close to \(e\). Hence, the horizontal asymptote for \(f(x)\) on this interval is the constant value \(y = e\). The asymptote is \(y = 2.71828\).
05

Writing the Approximate Equation of the Asymptote

The horizontal asymptote is \(y = e\). To approximate, we can express this as \(y = 2.72\) for simplicity and practical calculation within the given range \((0, 200]\).

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Key Concepts

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

Exponential Function
Exponential functions are a class of mathematical expressions characterized by a constant base raised to a variable exponent. These functions have distinct properties that cause them to grow or decay at consistent rates. In the case of the function \(f(x) = \left(1 + \frac{1}{x}\right)^{x}\), we see a special form that resembles an exponential function when examined as \(x\) approaches larger values.
Exponential functions exhibit rapid growth as the variable increases.
Common features include:
  • A constant rate of change, meaning the function can model growth like population growth or radioactive decay.
  • An asymptotic approach to zero, infinity, or a constant value, depending on the exponent’s rate and direction.
  • The base of the exponent, often noted as \(b\), is typically greater than zero and not equal to one.
The form \(\left(1 + \frac{1}{x}\right)^{x}\) interestingly approaches the natural base \(e\) as \(x\) increases, acting similarly to how exponential growth behaves with a fixed rate but becoming asymptotic because of its unique structure.
Euler's Number
Euler's number, commonly denoted by \(e\), is a fundamental mathematical constant approximately equal to 2.71828. Its importance in mathematics arises from its pervasive presence in the study of calculus, complex numbers, and various branches of mathematics involving growth processes.
Euler's number is the base of the natural logarithm. This makes it a crucial element in many exponential functions, particularly in equations involving continuous growth or decay. The function \(\left(1 + \frac{1}{x}\right)^{x}\) beautifully demonstrates Euler's number when \(x\) grows large. This form guides us to understand the behavior of exponential relationships over increasing intervals.
Key points about \(e\) include:
  • It can be seen as the limiting value of \(\left(1 + \frac{1}{n}\right)^{n}\) as \(n\) goes to infinity.
  • It's utilized in calculating compound interest, population dynamics, and in solving differential equations.
  • Because it is irrational, \(e\) has an infinite, non-repeating decimal expansion.
In our function, \(f(x)\), you observe \(e\)'s influence as \(x\) increases, pushing the output of the function towards this defining constant.
Function Behavior
Understanding function behavior reveals how inputs relate to outputs and the wider implications of a function's graph. For the function \(f(x) = \left(1 + \frac{1}{x}\right)^{x}\), its behavior as \(x\) approaches larger values is essential for identifying asymptotic tendencies.
As \(x\) increases, the fraction \(\frac{1}{x}\) decreases, leading to an interesting effect on \(f(x)\). The base \(1 + \frac{1}{x}\) approaches 1, yet the exponent \(x\) increases, causing \(f(x)\) to stabilize near \(e\). This gradual settling forms a horizontal asymptote, a line that the curve will approach but never actually reach as \(x\) grows.
Important concepts in function behavior include:
  • The limit: As \(x\) heads towards infinity, \(\left(1 + \frac{1}{x}\right)^{x}\) moves closer to \(e\).
  • Graph smoothness and continuity: Over the interval \((0, 200]\), \(f(x)\) shows a reflective trend of steady growth becoming consistent at higher \(x\) values, nearing \(e\).
  • Asymptotic analysis: Providing insights into the long-run behavior of the function, allowing predictions of behavior without plotting every point.
In conclusion, the behavior observed provides a practical example of the subtle interaction between a function's terms leading to the crucial concept of horizontal asymptotes.

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Most popular questions from this chapter

Dissolving salt in water If 10 grams of salt is added to a quantity of water, then the amount \(q(t)\) that is undissolved after \(t\) minutes is given by \(q(t)=10\left(\frac{4}{5}\right)^{t}\). Sketch a graph that shows the value \(q(t)\) at any time from \(t=0\) to \(t=10\)

Change \(f(x)=1000(1.05)^{x}\) to an exponential function with base \(e\) and approximate the growth rate of \(f\).

A drug is eliminated from the body through urine. Suppose that for an initial dose of 10 milligrams, the amount \(A(t)\) in the body \(t\) hours later is given by \(A(t)=10(0.8)^{t}\) (a) Estimate the amount of the drug in the body 8 hours after the initial dose. (b) What percentage of the drug still in the body is eliminated each hour?

(a) Graph \(f\) using a graphing utility. (b) Sketch the graph of \(g\) by taking the reciprocals of \(y\) -coordinates in (a), without using a graphing utility. $$f(x)=\frac{e^{x}+e^{-x}}{2} ; \quad g(x)=\frac{2}{e^{x}+e^{-x}}$$

Studies relating serum cholesterol level to coronary heart disease suggest that a risk factor is the ratio \(x\) of the total amount \(C\) of cholesterol in the blood to the amount \(H\) of high-density lipoprotein cholesterol in the blood. For a female, the lifetime risk \(R\) of having a heart attack can be approximated by the formula $$R=2.07 \ln x-2.04 \quad \text { provided } \quad 0 \leq R \leq 1$$ For example, if \(R=0.65,\) then there is a \(65 \%\) chance that a woman will have a heart attack over an average lifetime. (a) Calculate \(R\) for a female with \(C=242\) and \(H=78\) (b) Graphically estimate \(x\) when the risk is \(75 \%\)

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