Problem 2
Question
Which of the following functions grow faster than \(e^{x}\) as \(x \rightarrow \infty ?\) Which grow at the same rate as \(e^{x}\) ? Which grow slower? $$ \begin{array}{ll}{\text { a. } 10 x^{4}+30 x+1} & {\text { b. } x \ln x-x} \\\ {\text { c. } \sqrt{1+x^{4}}} & {\text { d. }(5 / 2)^{x}} \\ {\text { e. }} {e^{-x}} & {\text { f. } x e^{x}} \\ {\text { g. }} e^{\cos x}& {\text { h. } e^{x-1}}\end{array} $$
Step-by-Step Solution
Verified Answer
Functions d and f grow faster than \(e^x\). Function h grows at the same rate. All others grow slower.
1Step 1: Understanding Exponential Growth
The exponential function \(e^x\) grows very rapidly as \(x\to\infty\). Therefore, we need to compare the growth of other functions to \(e^x\). A function grows faster than \(e^x\) if, as \(x\rightarrow\infty\), the ratio \(f(x)/e^x\) tends to infinity. If this ratio tends to zero, it grows slower, and if it tends to a constant, it grows at the same rate.
2Step 2: Analyze Function (a)
Function: \(f(x) = 10x^4 + 30x + 1\). As \(x\to\infty\), polynomial functions grow slower than exponential functions. \[ \lim_{x\to\infty} \frac{10x^4 + 30x + 1}{e^x} = 0 \]. Thus, it grows slower than \(e^x\).
3Step 3: Analyze Function (b)
Function: \(f(x) = x \ln x - x\). Both \(x \ln x\) and \(x\) grow slower than \(e^x\) as \(x\to\infty\). \[ \lim_{x\to\infty} \frac{x \ln x - x}{e^x} = 0 \]. So, it grows slower than \(e^x\).
4Step 4: Analyze Function (c)
Function: \(f(x) = \sqrt{1+x^4}\). As \(x\to\infty\), this behaves like \(x^2\) because \(\sqrt{1+x^4}\approx x^2\). \[ \lim_{x\to\infty} \frac{\sqrt{1+x^4}}{e^x} = 0 \]. Thus, it grows slower than \(e^x\).
5Step 5: Analyze Function (d)
Function: \(f(x) = (5/2)^x\). This is an exponential function but with a base greater than \(e\), since \(5/2 > e\). As \(x\to\infty\), \[ \lim_{x\to\infty} \frac{(5/2)^x}{e^x} = \infty \]. Therefore, it grows faster than \(e^x\).
6Step 6: Analyze Function (e)
Function: \(f(x) = e^{-x}\). This is equivalent to \(1/e^x\), which clearly grows much slower because it approaches 0 as \(x\to\infty\). \[ \lim_{x\to\infty} \frac{e^{-x}}{e^x} = \lim_{x\to\infty} e^{-2x} = 0 \]. Thus, it grows slower than \(e^x\).
7Step 7: Analyze Function (f)
Function: \(f(x) = xe^x\). Here, \[ \lim_{x\to\infty} \frac{xe^x}{e^x} = \lim_{x\to\infty} x = \infty \]. So, this grows faster than \(e^x\) because the \(x\) increases the rate.
8Step 8: Analyze Function (g)
Function: \(f(x) = e^{\cos x}\). Since \(\cos x\) oscillates between -1 and 1, \(e^{\cos x}\) oscillates but remains bounded, so \[ \lim_{x\to\infty} \frac{e^{\cos x}}{e^x} = \lim_{x\to\infty} e^{\cos x - x} = 0 \]. Thus, it grows slower than \(e^x\).
9Step 9: Analyze Function (h)
Function: \(f(x) = e^{x-1}\). This is basically \(e^x/e = (1/e)e^x\). Thus, \[ \lim_{x\to\infty} \frac{e^{x-1}}{e^x} = \lim_{x\to\infty} \frac{e^{x}}{e \cdot e^x} = \frac{1}{e} \], which is constant. So, it grows at the same rate as \(e^x\).
Key Concepts
Growth Rate ComparisonPolynomial vs Exponential FunctionsFunction Limit AnalysisLogarithmic Function Growth
Growth Rate Comparison
When trying to understand how different mathematical functions grow with respect to each other, especially as a variable approaches infinity, it is essential to compare their growth rates. A powerful tool for this is the concept of limits, which help determine the behavior of functions as they extend towards infinity.
Consider an example with the exponential function \(e^x\). To determine if another function \(f(x)\) grows faster, equally, or slower than \(e^x\), we evaluate the limit \(\lim_{{x \to \infty}} \frac{f(x)}{e^x}\):
Consider an example with the exponential function \(e^x\). To determine if another function \(f(x)\) grows faster, equally, or slower than \(e^x\), we evaluate the limit \(\lim_{{x \to \infty}} \frac{f(x)}{e^x}\):
- If this limit is infinity, then \(f(x)\) grows faster than \(e^x\).
- If it results in zero, \(f(x)\) grows slower.
- If the limit equals a constant, both functions grow at the same rate.
Polynomial vs Exponential Functions
Polynomial and exponential functions behave very differently as they grow. When comparing the growth rates, exponential functions often outpace polynomial functions dramatically. This is because exponential functions have a constant base raised to the power of the variable, leading to rapid growth.
For instance, a function like \(10x^4\) (which is a polynomial), when compared to an exponential function like \(e^x\), shows a remarkable difference as \(x\) becomes very large:
- The limit \(\lim_{{x \to \infty}} \frac{10x^4}{e^x} = 0\) illustrates that polynomial functions, no matter the degree, grow slower than exponential ones. This is why exponential growth models, like population or finance models, are often used to depict fast-growing situations.
For instance, a function like \(10x^4\) (which is a polynomial), when compared to an exponential function like \(e^x\), shows a remarkable difference as \(x\) becomes very large:
- The limit \(\lim_{{x \to \infty}} \frac{10x^4}{e^x} = 0\) illustrates that polynomial functions, no matter the degree, grow slower than exponential ones. This is why exponential growth models, like population or finance models, are often used to depict fast-growing situations.
Function Limit Analysis
The limit analysis of functions helps us predict their behavior for very large inputs, specifically as \(x\to \infty\). This analysis is crucial for understanding the growth pattern of various functions in comparison to one another.
For analyzing how fast or slow functions grow, you perform calculations such as:
- \(\lim_{{x \to \infty}} \frac{(5/2)^x}{e^x} = \infty\) implies that the base \(5/2\) which is greater than \(e\) grows much faster than \(e^x\).
- Conversely, functions such as \(e^{-x}\), where \(\lim_{{x \to \infty}} \frac{e^{-x}}{e^x} = 0\), show how functions decay or grow slower over time.
Such limit evaluations help in defining the speed and magnitude of growth and are widely used in calculus and applied mathematics to evaluate real-world scenarios.
For analyzing how fast or slow functions grow, you perform calculations such as:
- \(\lim_{{x \to \infty}} \frac{(5/2)^x}{e^x} = \infty\) implies that the base \(5/2\) which is greater than \(e\) grows much faster than \(e^x\).
- Conversely, functions such as \(e^{-x}\), where \(\lim_{{x \to \infty}} \frac{e^{-x}}{e^x} = 0\), show how functions decay or grow slower over time.
Such limit evaluations help in defining the speed and magnitude of growth and are widely used in calculus and applied mathematics to evaluate real-world scenarios.
Logarithmic Function Growth
Logarithmic functions grow slower compared to polynomial and exponential functions. The function \(\ln x\) grows very slowly and is often utilized in scenarios where rapid growth is unsustainable or impractical.
In terms of comparing with an exponential function, a logarithmic function like \(x \ln x\) behaves as follows:
In terms of comparing with an exponential function, a logarithmic function like \(x \ln x\) behaves as follows:
- The limit \(\lim_{{x \to \infty}} \frac{x \ln x - x}{e^x} = 0\) explains how both the terms within grow much slower than \(e^x\).
Other exercises in this chapter
Problem 1
solve for \(t.\) \begin{equation}\quad \text { a. }e^{-0.3 t}=27 \quad \text { b. } e^{k t}=\frac{1}{2} \quad \text { c. } e^{(\ln 0.2) t}=0.4\end{equation}
View solution Problem 1
Express the following logarithms in terms of \(\ln 2\) and \(\ln 3\) $$\begin{array}{lll}{\text { a. } \ln 0.75} & {\text { b. } \ln (4 / 9)} & {\text { c. } \l
View solution Problem 2
Each of Exercises \(1-4\) gives a value of sinh \(x\) or cosh \(x .\) Use the definitions and the identity \(\cosh ^{2} x-\sinh ^{2} x=1\) to find the values of
View solution Problem 2
In Exercises \(1-4,\) show that each function \(y=f(x)\) is a solution of the accompanying differential equation. $$\begin{array}{l}{y^{\prime}=y^{2}} \\ {\text
View solution