Problem 25 - College Algebra

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

Local behavior: increasing everywhere. End behavior: \(f(x)\to\frac{1}{2}\) as \(x\to\pm\infty\). Vertical asymptote at \(x=-\frac{1}{2}\).

Step 1: Identify Critical Points

To understand the local behavior of the function, identify critical points by taking the derivative. First, simplifiy the function: \(f(x) = \frac{x}{2x+1}\). Using quotient rule where \(u = x\) and \(v = 2x+1\), the derivative \(f'(x)\) is calculated using \(f'(x) = \frac{v\cdot u' - u \cdot v'}{v^2}\). Therefore, \(f'(x) = \frac{(2x+1)(1) - x(2)}{(2x+1)^2}\). Simplify to get \(f'(x) = \frac{2x+1-2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}\). Since \(f'(x)\) is never zero, and the denominator is never negative, no critical points exist.

Step 2: Analyze Local Behavior

Based on the first derivative test, since \(f'(x)\) has a positive value for all \(x\), the function is increasing everywhere on its domain. Thus, \(f(x)\) has no local maxima or minima.

Step 3: Determine the Domain

The function \(f(x) = \frac{x}{2x+1}\) is defined for all real numbers x except when the denominator equals zero. Set the denominator equal to zero: \(2x+1 = 0\), solving gives \(x = -\frac{1}{2}\). Therefore, the domain is \(x \in (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)\).

Step 4: Analyze End Behavior for Large \(x\)

To find end behavior, evaluate the limits as \(x\) approaches positive and negative infinity. As \(x \to \infty\), the dominant terms in the numerator and denominator are both \(x\) and \(2x\) respectively. Hence, \(f(x) \to \frac{x}{2x} = \frac{1}{2}\). Thus, as \(x\to \infty\), \(f(x)\to\frac{1}{2}\). As \(x \to -\infty\), \(f(x)\to \frac{1}{2}\) because the major terms still dominate.

Step 5: Analyze Behavior Near Undefined Point

Evaluate the behavior as \(x\) approaches the undefined value of \(-\frac{1}{2}\). Consider the limits from the left and the right. As \(x \to -\frac{1}{2}^{-}\), the denominator approaches 0 from the negative side, thus, \(f(x)\to -\infty\). As \(x \to -\frac{1}{2}^{+}\), the denominator approaches 0 from the positive side, thus \(f(x)\to \infty\). There is a vertical asymptote at \(x = -\frac{1}{2}\).

Step 6: Summarize Behaviors

Local behavior: Function increases everywhere due to a positive derivative with no critical points. End behavior: As \(x\to\infty\) and \(x\to-\infty\), \(f(x)\to\frac{1}{2}\). Near \(x = -\frac{1}{2}\), \(f(x)\) approaches negative and positive infinity indicating a vertical asymptote.

Key Concepts

Critical Points

Critical points of a function are where its derivative is zero or undefined, helping us to identify where a function might have local maxima, minima, or saddle points. For the function \( f(x) = \frac{x}{2x+1} \), we first need to find its derivative to locate any critical points. Using the quotient rule, which states \( f'(x) = \frac{v u' - u v'}{v^2} \), we find that the derivative simplifies to \( f'(x) = \frac{1}{(2x+1)^2} \).

Here, the numerator is always 1, meaning \( f'(x) \) never equals zero, and the denominator is always positive since \( 2x+1 \) squared cannot be negative. Thus, there are no points where the derivative equals zero or is undefined, indicating that this function has no critical points.

Remember, the absence of critical points suggests the function either increases or decreases monotonically over its domain without any local extrema.

Vertical Asymptote

A vertical asymptote occurs when the function's value grows infinitely large or infinitely small as it approaches some \( x \) value. For the function \( f(x) = \frac{x}{2x+1} \), a vertical asymptote occurs where the denominator is zero, because division by zero is undefined.

For this function, setting the denominator \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \). This is where the vertical asymptote occurs. As \( x \to -\frac{1}{2}^{-} \), the value of \( f(x) \to -\infty \), and as \( x \to -\frac{1}{2}^{+} \), \( f(x) \to \infty \).

This asymptotic behavior suggests that as \( x \) approaches \( -\frac{1}{2} \) from either side, the function shoots up in opposite directions, a hallmark of vertical asymptotes.

Domain of a Function

The domain of a function determines all the possible inputs (x-values) for which the function is defined. The function \( f(x) = \frac{x}{2x+1} \) is a rational function, meaning it has a polynomial in the numerator and denominator.

The domain is all real numbers except where the denominator equals zero, as division by zero is undefined. Solving \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \). Thus, the domain excludes \( x = -\frac{1}{2} \).

We can express the domain of the function as all real numbers except \( x = -\frac{1}{2} \), formally written as \( x \in (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty) \). This clearly delineates the region where the function is valid and helps avoid undefined behavior at \( x = -\frac{1}{2} \).

First Derivative Test

The First Derivative Test is a method for determining the local maxima and minima of a function. If the derivative is positive over an interval, the function is increasing; if negative, it is decreasing. This test analyzes the sign change of the derivative before and after a critical point.

For the function \( f(x) = \frac{x}{2x+1} \), since we determined the derivative \( f'(x) = \frac{1}{(2x+1)^2} \) is always positive, the function is consistently increasing over its entire domain. There is no change in the sign of the derivative because \( f'(x) \) never approaches zero or becomes negative.

This means the function has no local maxima or minima, as it exhibits a uniform increasing behavior over all x-values where it is defined. This consistent increase underlines the lack of critical points and extrema.