Question

In Exercises, find the critical numbers and the open intervals on which the function is increasing or decreasing. (Hint: Check for discontinuities.) Sketch the graph of the function. $$ f(x)=\frac{x}{x+1} $$

Short answer

The function \( f(x) = \frac{x}{x+1} \) has no critical numbers, a discontinuity at \( x = -1 \), and is increasing for all other values of x.

Step-by-step solution

Find the Derivative

To find where the function increases or decreases, find the derivative of the function. The derivative of \( f(x) = \frac{x}{x+1} \) is \( f'(x) = \frac{1}{(x+1)^2} \).

Find the Critical Numbers

Set the derivative equal to zero and solve for x to find critical numbers. However, because the fraction \( \frac{1}{(x+1)^2} \) cannot be zero, there are no critical numbers from setting the derivative equal to zero. Also check for any numbers that make the derivative undefined. This occurs when \( x+1 = 0 \), or \( x = -1 \). Therefore, the function has one critical number, \( x = -1 \).

Find the Intervals of Increase and Decrease

Use the critical number to divide the domain of the function into intervals. Test numbers in each interval by plugging them into the derivative. If the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing. The function is increasing for \( x < -1 \) and \( x > -1 \) since the derivative is always positive. The function has a discontinuity at \( x = -1 \).

Sketch the graph

Based on the analysis, the graph of the function decreases as it approaches \( x = -1 \) from the left, has a discontinuity at \( x = -1 \), and then increases as x moves away from \( x = -1 \) to the right. This confirms that \( x = -1 \) is the only discontinuity and the function is increasing for all other x-values.

Key concepts

Understanding the Derivative

To analyze how a function behaves, the derivative acts as an important tool. It helps us determine where a function is increasing or decreasing.
For the function \( f(x) = \frac{x}{x+1} \), we calculate its derivative to gain insights into its nature. The derivative is found using the quotient rule. This result is \( f'(x) = \frac{1}{(x+1)^2} \).
This derivative tells us how the function's rate of change behaves. Notice that \( \frac{1}{(x+1)^2} \) is always positive when defined, indicating an increasing function. However, we also need to consider points where the derivative might not be defined.

Increasing and Decreasing Intervals

The intervals on which a function increases or decreases can be determined through the sign of its derivative. Here, \( f'(x) = \frac{1}{(x+1)^2} \) is strictly positive and never zero in its domain where it is defined, which suggests the function is increasing in these regions.
Next, identify where the derivative is not defined due to division by zero. In this case, \( x = -1 \) makes \( (x+1)^2 = 0 \). Hence, the function is not defined here, splitting the number line into \( (\infty, -1) \) and \( (-1, \infty) \). In both intervals, the derivative is positive, implying the function increases to the left and right of \( x = -1 \).

Identifying Discontinuities

Discontinuities in a function occur where the function is not defined or lacks a limit. For \( f(x) = \frac{x}{x+1} \), the point \( x = -1 \) causes the denominator to become zero, leading to a discontinuity.
At this point, the function does not exist, and so, cannot be continued smoothly. This type of discontinuity is classified as a vertical asymptote, meaning as you approach \( x = -1 \) from either side, the function approaches infinity or drops to negative infinity. This break in the graph must be noted, as it separates the intervals where increases or decreases occur.

Graphing the Function

When sketching \( f(x) = \frac{x}{x+1} \), the key is to reflect on the derivative and detected discontinuities. The graph shows increasing behavior, consistent before and after \( x = -1 \).
However, due to the vertical asymptote at \( x = -1 \), the graph shows a significant break here. Before this point, the function rises from negative infinity towards \( x = -1 \), and post \( x = -1 \), it rises from negative infinity again.
A function that appears to drop and then rise sharply can be misleading without awareness of the vertical asymptote. Including this feature in your graph ensures a comprehensive understanding of \( f(x) \).