Problem 18
Question
Show that \(\sqrt{x^{4}+x}\) and \(\sqrt{x^{4}-x^{3}}\) grow at the same rate as \(x \rightarrow \infty\) by showing that they both grow at the same rate as \(x^{2}\) as \(x \rightarrow \infty\)
Step-by-Step Solution
Verified Answer
Both functions grow at the rate of \(x^2\) as \(x \to \infty\).
1Step 1: Simplify the Expression for Large x
Consider \(\sqrt{x^4 + x}\) as \(x \to \infty\), the dominant term inside the square root is \(x^4\). So, \(\sqrt{x^4 + x}\) simplifies to \(\sqrt{x^4(1 + \frac{x}{x^4})} = x^2\sqrt{1 + \frac{1}{x^3}}\). As \(x \to \infty\), the term \(\frac{1}{x^3}\) approaches zero, hence it simplifies to \(x^2\).
2Step 2: Simplify the Other Expression for Large x
Now consider \(\sqrt{x^4 - x^3}\). Similarly, the dominant term \(x^4\) gives \(\sqrt{x^4(1 - \frac{x^3}{x^4})} = x^2\sqrt{1 - \frac{1}{x}}\). As \(x \to \infty\), \(\frac{1}{x}\) tends to zero, leaving us with \(x^2\).
3Step 3: Comparison to x^2
Both expressions simplify to \(x^2\) as \(x \to \infty\). This means that both \(\sqrt{x^4 + x}\) and \(\sqrt{x^4 - x^3}\) grow at the same rate as \(x^2\), confirming that they grow at the same rate as each other as \(x \to \infty\).
Key Concepts
Dominant TermSquare Root SimplificationGrowth Rate Comparison
Dominant Term
When working with complex functions, especially for very large values of the variable, it's useful to determine the dominant term inside an expression. The dominant term is the part of the expression that grows the fastest and has the greatest impact on the function's behavior for large values. In this context, it is especially relevant for simplification and asymptotic analysis. Consider the expression \( \sqrt{x^4 + x} \). As \( x \to \infty \), while both terms inside the square root, \( x^4 \) and \( x \), grow, \( x^4 \) grows much faster than \( x \). Therefore, \( x^4 \) is the dominant term.
- The expression simplifies to \( \sqrt{x^4(1 + \frac{1}{x^3})} \) by factoring out \( x^4 \).
- As \( x \) becomes very large, \( \frac{1}{x^3} \) becomes negligible, simplifying the square root to \( x^2 \).
- It simplifies to \( \sqrt{x^4(1 - \frac{1}{x})} \).
- Since \( \frac{1}{x} \to 0 \), the square root approaches \( x^2 \).
Square Root Simplification
Square root simplification is a critical tool used in analyzing functions, especially within asymptotic analysis. By simplifying the square root, we can extract useful information about the growth behavior of functions as variables become large, typically represented by \( x \to \infty \). This step focuses on reducing complex expressions to forms that are easier to compare and interpret.Given the expression \( \sqrt{x^4 + x} \), the simplification process involves understanding that the terms within the square root can be split.
- For \( x^4 + x \), factoring out \( x^4 \) gives \( \sqrt{x^4(1 + \frac{x}{x^4})} = x^2\sqrt{1 + \frac{1}{x^3}} \).
- As \( x \) increases, \( \frac{1}{x^3} \to 0 \), making the radical approximate \( x^2 \).
- It simplifies as \( \sqrt{x^4(1 - \frac{x^3}{x^4})} = x^2\sqrt{1 - \frac{1}{x}} \).
- When \( x \) is large, \( \frac{1}{x} \) becomes negligible, leading again to \( x^2 \).
Growth Rate Comparison
In the realm of asymptotic analysis, comparing growth rates is essential for understanding how different functions behave as variables tend toward infinity. The objective is to determine whether two or more functions grow at the same speed by observing their simplifications and dominant terms.For functions \( \sqrt{x^4 + x} \) and \( \sqrt{x^4 - x^3} \), after identifying the dominant term and simplifying the square roots, we noted that both expressions simplify to \( x^2 \) as \( x \to \infty \). Therefore, even if their full expressions differ, their growth rates are identical.
- The simplification of each function to \( x^2 \) establishes that, asymptotically, they can be treated equivalently in terms of growth rate.
- Thus, both \( \sqrt{x^4 + x} \) and \( \sqrt{x^4 - x^3} \) grow at the same rate as \( x^2 \).
Other exercises in this chapter
Problem 17
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=\cos \left(e^{-\theta^{2}}\right)\end{equation}
View solution Problem 17
In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=t(\ln t)^{2} $$
View solution Problem 18
In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\ln (\cosh z)$$
View solution Problem 18
Solve the differential equations in Exercises \(9-22\) $$\frac{d y}{d x}=\frac{e^{2 x-y}}{e^{x+y}}$$
View solution