Problem 479 - Precalculus

Question

For the following exercises, describe the local and end behavior of the functions. $$f(x)=\frac{x}{2 x+1}$$

Step-by-Step Solution

Answer

The function is increasing, has a horizontal asymptote at \( y = \frac{1}{2} \), and a vertical asymptote at \( x = -\frac{1}{2} \).

Step 1: Identify the Local Behavior

The local behavior of the function can be analyzed by examining the derivative. Since we are given the function \( f(x) = \frac{x}{2x+1} \), let's find its derivative to determine increasing or decreasing behavior and locate any critical points. Use the quotient rule here: if \( u = x \) and \( v = 2x + 1 \), then \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = 2 \). The derivative \( f'(x) \) is given by:\[f'(x) = \frac{(2x+1)(1) - x(2)}{(2x+1)^2} = \frac{2x + 1 - 2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}.\]Since \( f'(x) > 0 \) for all \( x \) in the domain, the function is strictly increasing.

Step 2: Determine the Horizontal Asymptote

To find the end behavior, examine \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \). For the function \( f(x) = \frac{x}{2x+1} \), divide the numerator and the denominator by \( x \):\[\lim_{x \to \infty} \frac{x}{2x+1} = \lim_{x \to \infty} \frac{1}{2 + \frac{1}{x}} = \frac{1}{2}.\]Similarly,\[\lim_{x \to -\infty} \frac{x}{2x+1} = \frac{1}{2}.\]Thus, the function approaches the horizontal asymptote \( y = \frac{1}{2} \) as \( x \to \pm \infty \).

Step 3: Address the Vertical Asymptote

Vertical asymptotes occur where the denominator is zero, and the numerator is not zero. Set \( 2x+1 = 0 \) to find the vertical asymptote. Solving this gives:\[2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}.\]This means the function has a vertical asymptote at \( x = -\frac{1}{2} \).

Step 4: Summary of Behaviors

Summarizing the behavior:- Local Behavior: Since the derivative \( f'(x) > 0 \) for all \( x eq -\frac{1}{2} \), \( f(x) \) is increasing everywhere it is defined.- End Behavior: As \( x \to \, \pm \infty \), \( f(x) \to \frac{1}{2} \).- Vertical Asymptote: There is a vertical asymptote at \( x = -\frac{1}{2} \).

Key Concepts

Derivative Analysis

Analyzing the derivative of a function is crucial in understanding its local behavior. The derivative, denoted as \( f'(x) \), tells us how the function \( f(x) \) changes around any given point. In simpler terms, it indicates whether the function is increasing or decreasing. Let's consider the derivative of the given function \( f(x) = \frac{x}{2x+1} \).To find the derivative, we used the quotient rule, which is appropriate for functions where one function is divided by another. Here, we identified the functions \( u = x \) and \( v = 2x + 1 \), resulting in \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = 2 \). Using these, the quotient rule gives:\[f'(x) = \frac{(2x+1)\cdot1 - x\cdot2}{(2x+1)^2} = \frac{1}{(2x+1)^2}.\]An interesting fact about this derivative is that \( f'(x) > 0 \) for all \( x \) within the domain of \( f(x) \). This means the function is always increasing, aside from the point where the function is undefined due to the vertical asymptote.

Horizontal Asymptote

A horizontal asymptote describes the behavior of a function as \( x \) approaches infinity or negative infinity. It is a horizontal line that the graph of the function approaches but never touches. For the function \( f(x) = \frac{x}{2x+1} \), determining the horizontal asymptote involves evaluating the limits as \( x \to \pm \infty \).By simplifying the function as \( x \to \infty \) or \( x \to -\infty \), we divide the numerator and the denominator by \( x \):\[\lim_{x \to \infty} \frac{x}{2x+1} = \lim_{x \to \infty} \frac{1}{2 + \frac{1}{x}} = \frac{1}{2}.\]This calculation shows that the function approaches the value \( y = \frac{1}{2} \) as \( x \to \pm \infty \). Thus, the horizontal asymptote of the function is at \( y = \frac{1}{2} \), signifying that, no matter how large or small \( x \) becomes, the function's value will get closer and closer to \( 0.5 \), but never quite reach it.

Vertical Asymptote

Vertical asymptotes occur when the function approaches a particular value of \( x \), and the function’s output heads towards infinity, either positively or negatively. In simpler terms, this means that the graph of the function will shoot up or down, but never cross that line.For the function \( f(x) = \frac{x}{2x+1} \), a vertical asymptote occurs where the denominator equals zero while the numerator does not. To find where this happens, set the denominator \( 2x + 1 \) equal to zero:\[2x + 1 = 0 \x = -\frac{1}{2}.\]Therefore, there is a vertical asymptote at \( x = -\frac{1}{2} \). As the function value nears \( x = -\frac{1}{2} \) from either side, the value of \( f(x) \) heads towards infinity or negative infinity, making it an essential feature in understanding the behavior and limitations of the function.