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Order the following functions from slowest growing to fastest growing as \(x \rightarrow \infty\) a. \(2^{x}\) b. \(x^{2}\) c. \((\ln 2)^{x}\) d. \(e^{x}\)

Short Answer

Expert verified
Order: \((\ln 2)^{x}\), \(x^{2}\), \(2^{x}\), \(e^{x}\).

Step by step solution

01

Understand the Concept

Asymptotic growth is used to compare how fast different functions grow as the input, in this case, \(x\), approaches infinity. The idea is to compare the rate at which each function's value increases.
02

Analyze the Exponential Functions

In the list, both \(2^{x}\) and \(e^{x}\) are exponential functions. Among exponential functions, the base with the larger value grows faster. Here, \(e\) is approximately 2.718, which means \(e^{x}\) grows faster than \(2^{x}\).
03

Analyze the Power Function

The function \(x^{2}\) is a polynomial function. As a general rule, polynomial functions like \(x^{2}\) grow slower than exponential functions (e.g., \(2^{x}\) or \(e^{x}\)).
04

Analyze the Function with a Logarithm Base

The function \((\ln 2)^{x}\) is another kind of exponential function, where the base is the natural logarithm of 2, which is approximately 0.693. This base is less than 1, leading to a decay in growth, meaning it grows slower than any positive power of \(x\).
05

Order the Functions

Based on the analysis:1. \((\ln 2)^{x}\) grows the slowest because it decays as \(x\) increases.2. \(x^{2}\) grows faster than \((\ln 2)^{x}\) because it's a polynomial.3. \(2^{x}\) grows faster than \(x^{2}\) as it's an exponential function with a base greater than 1.4. \(e^{x}\) grows the fastest as it is an exponential function with a base greater than 2.

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Key Concepts

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

Exponential Functions
Exponential functions are characterized by a constant base raised to a variable exponent, typically in the form of \( a^x \), where \( a \) is a positive constant. These functions grow at an extremely rapid rate compared to other types of functions. The key characteristic of exponential growth is that as \( x \) increases, the function's values multiply rather than add. This means if \( x \) increases by 1, the value of the function is multiplied by \( a \), making them grow faster and faster as \( x \) becomes larger.
The value of the base \( a \) is crucial in determining how fast the function will grow.
Here are some key pointers about exponential functions:
  • If \( a > 1 \), the function shows rapid growth as \( x \) increases.
  • If \( 0 < a < 1 \), the function decays, meaning its values get closer to zero as \( x \) increases.
  • Common bases include \( e \), approximately 2.718, and integers greater than 1, such as 2 or 10.
In our example, both \( 2^x \) and \( e^x \) are exponential. \( e^x \) grows faster than \( 2^x \) because \( e \) is greater than 2, showing the profound impact the base value has on the growth rate.
Polynomial Functions
Polynomial functions are expressions involving a sum of powers of variables. These functions typically appear as \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \), where the powers of \( x \) are non-negative integers. The degree, \( n \), indicates the highest power of the polynomial and determines the basic shape and growth of the graph.
For instance, a polynomial function like \( x^2 \) consists of a monomial with a power of 2.
Here are some facts about polynomial functions:
  • As \( x \) tends towards infinity, the leading term (the term with the highest power) dominates the growth behavior.
  • Polynomial growth is generally slower than exponential growth; for example, \( x^n \) grows more slowly than \( a^x \) (where \( a > 1 \)).
  • Unlike exponential functions, adding a constant or coefficient has a more straightforward, stable impact on growth.
In our exercise, \( x^2 \) grows slower than any exponential function like \( 2^x \), illustrating that exponential increases outpace polynomial increases as \( x \) becomes very large.
Growth Rate Comparison
Comparing the growth of different functions helps understand their long-term behavior. The core idea behind asymptotic growth is to determine which function will eventually outgrow the others as \( x \) approaches infinity.
The rate of growth can be categorized generally as decay (very slow growth), polynomial growth, and exponential growth as follows:
  • Decay: Occurs when the function values decrease as \( x \) increases, such as when the base of an exponential function is between 0 and 1, like \( (\ln 2)^x \).
  • Polynomial Growth: Functions like \( x^2 \) grow in a steady manner but will eventually be outpaced by exponential functions.
  • Exponential Growth: This is very rapid and functions such as \( 2^x \) and \( e^x \) grow quicker than any polynomial function. Of these, \( e^x \) grows the fastest in our example.
In the given exercise, we saw: \( (\ln 2)^x \) grows the slowest (decay), followed by the polynomial \( x^2 \), then the exponential \( 2^x \), and fastest is \( e^x \). Understanding this hierarchy in growth rates helps in fields like computer science, economics, and natural sciences, where predicting long-term behavior is crucial.

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Most popular questions from this chapter

a. Graph \(f(x)=(\sin x)^{x}\) on the interval \(0 \leq x \leq \pi .\) What value would you assign to \(f\) to make it continuous at \(x=0 ?\) b. Verify your conclusion in part (a) by finding \(\lim _{x \rightarrow 0^{+}} f(x)\) with l'Hôpital's Rule. c. Returning to the graph, estimate the maximum value of \(f\) on \([0, \pi] .\) About where is max \(f\) taken on? d. Sharpen your estimate in part (c) by graphing \(f^{\prime}\) in the same window to see where its graph crosses the \(x\) -axis. To simplify your work, you might want to delete the exponential factor from the expression for \(f^{\prime}\) and graph just the factor that has a zero.

Estimate the value of $$\lim _{x \rightarrow 1} \frac{2 x^{2}-(3 x+1) \sqrt{x}+2}{x-1}$$ by graphing. Then confirm your estimate with I'Hôpital's Rule.

Skydiving If a body of mass \(m\) falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body's velocity \(t\) s into the fall satisfies the differential equation $$m \frac{d v}{d t}=m g-k v^{2}$$ where \(k\) is a constant that depends on the body's aerodynamic properties and the density of the air. (We assume that the fall is short enough so that the variation in the air's density will not affect the outcome significantly.) a. Show that $$\boldsymbol{v}=\sqrt{\frac{m g}{k}} \tanh (\sqrt{\frac{g k}{m}} t)$$ satisfies the differential equation and the initial condition that \(v=0\) when \(t=0\) b. Find the body's limiting velocity, lim \(_{t \rightarrow \infty} v\) c. For a \(75 \mathrm{kg}\) skydiver \((m g=735 \mathrm{N}),\) with time in seconds and distance in meters, a typical value for \(k\) is \(0.235 .\) What is the diver's limiting velocity?

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=x^{\sin x}$$

Show that \(e^{x}\) grows faster as \(x \rightarrow \infty\) than any polynomial $$a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$$.

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