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Complete them to review topics relevant to the remaining exercises. Let \(f(x)=5 e^{x} .\) As \(x \rightarrow \infty, f(x) \rightarrow\)__________.

Short Answer

Expert verified
\(\lim_{{x \to \infty}} f(x) = \infty\).

Step by step solution

01

Explore the function

First, let's look closer to the given function \(f(x)=5e^x\). This is an exponential function where \(e\) is Euler's number, which is approximately equal to 2.71828.
02

Understand the behavior of exponential function.

The nature of exponential functions is such that the function will increase as the value of \(x\) increases. For \(f(x)=e^x\), as \(x\) approaches infinity, \(e^x\) also approaches infinity. Since the given function is \(f(x)=5e^x\), without any negative sign or term subtracting from it, as \(x\) approaches infinity, \(f(x)\) will also approach infinity.
03

Calculate the limit.

The limit of \(f(x)=5e^x\) as \(x\) approaches infinity is therefore infinity. We can write it formally as \(\lim_{{x\to \infty}} 5e^x = \infty\).

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Key Concepts

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

Limits
In mathematics, a limit describes the value that a function approaches as the input approaches a certain point. Limits are essential for understanding the behavior of functions, especially as the inputs grow larger or smaller. In our exercise, we use limits to determine what happens to the function \(f(x) = 5e^x\) as \(x\) gets very large, specifically as \(x\) approaches infinity. The notation
  • \(\lim_{{x\to \infty}} 5e^x\)
helps us express this idea concisely. When \(x\) becomes very large, the function \(5e^x\) keeps growing because \(e^x\) becomes larger and larger. Therefore, the limit of \(5e^x\) as \(x\) approaches infinity is infinity itself. This illustrates how limits offer a way to grasp what happens at the extremes of a function's domain, even when we can't compute an exact value.
Asymptotic Behavior
Asymptotic behavior gives us a deeper understanding of how functions act when inputs become extremely large or small. For exponential functions like \(f(x) = 5e^x\), their asymptotic behavior as \(x\) approaches infinity is critical. These functions grow very quickly, which we refer to as having an asymptote at infinity.An asymptote is a line that functions approach but never actually reach. In the case of \(5e^x\), as \(x\) grows, its asymptotic behavior shows a sharp increase toward infinity. Unlike other types of functions, like rational or polynomial functions that might level off or approach a horizontal line as \(x\) becomes large, exponential functions do not have horizontal asymptotes.Thus, we see that the asymptotic behavior of \(5e^x\) is steep and unbounded as \(x\) approaches infinity, highlighting the incredibly rapid growth rate of exponential functions.
Function Behavior
Function behavior refers to how a function acts or changes as \(x\) varies across its domain. For \(f(x) = 5e^x\), understanding its behavior means analyzing how it increases as \(x\) increases.Exponential functions are known for their rapid growth. As \(x\) grows, \(e^x\) increases dramatically, and being multiplied by 5 makes the growth of \(f(x)\) five times greater. There's no slowing down—the function's value keeps rising steadily without bounds.This critical behavior is evident from the graph of the function as well: it's a curve that rises sharply upwards as \(x\) extends to infinity. The function does not have peaks, valleys, or turning points; it just continuously increases. Moreover, such behavior is crucial in real-world applications, like modeling population growth or radioactive decay, where understanding how quickly a quantity is increasing or decreasing over time is important.

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Most popular questions from this chapter

Plutonium is a radioactive element that has a half-life of 24,360 years. The half-life of a radioactive substance is the time it takes for half of the substance to decay (which means the other half will still exist after that length of time). Find an exponential function of the form \(f(t)=A e^{k t}\) that gives the amount of plutonium left after \(t\) years if the initial amount of plutonium is 10 pounds. How long will it take for the plutonium to decay to 2 pounds?

Determine how long it takes for the given investment to double if \(r\) is the interest rate and the interest is compounded continuously. Assume that no withdrawals or further deposits are made. Initial amount: \(\$ 4000 ; r=5.75 \%\)

Solve the logarithmic equation and eliminate any extraneous solutions. If there are no solutions, so state. $$\log _{2}(x+5)=\log _{2}(x)+\log _{2}(x-3)$$

The cost of removing chemicals from drinking water depends on how much of the chemical can safcly be left behind in the water. The following table lists the annual removal costs for arsenic in terms of the concentration of arsenic in the drinking water. (Source: Environmental Protection Agency) $$\begin{array}{|c|c|}\hline\text { Arsenic Concentration } & \text { Annual Cost } \\\\\text { (micrograms per liter) } & \text { (millions of dollars) } \\\\\hline 3 & 645 \\\5 & 379 \\\10 & 166 \\\20 & 65\\\ \hline\end{array}$$ (a) Interpret the data in the table. What is the relation between the amount of arsenic left behind in the removal process and the annual cost? (One microgram is equal to \(10^{-6}\) gram.) (b) Make a scatter plot of the data and find the exponential function of the form \(C(x)=C a^{*}\) that best fits the data. Here, \(x\) is the arscnic concentration. (c) Why must \(a\) be less than 1 in your model? (d) Using your model, what is the annual cost to obtain an arsenic concentration of 12 micrograms per liter? (e) It would be best to have the smallest possible amount of arsenic in the drinking water, but the cost may be prohibitive. Use your model to calculate the annual cost of processing such that the concentration of arsenic is only 2 micrograms per liter of water. Interpret your result.

Evaluate the given quantity by referring to the function \(f\) given in the following table. $$\begin{array}{cc}x & f(x) \\\\-2 & 1 \\\\-1 & 2 \\\0 & 0 \\\1 & -1 \\\2 & -2\end{array}$$ $$f^{-1}(2)$$

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