Problem 25 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. $$ f(x)=\frac{1}{x-2} $$

Step-by-Step Solution

Answer

The function \(f(x) = \frac{1}{x - 2}\) is decreasing on the intervals \((-\infty, 2)\) and \((2, \infty)\), and there is no interval where the function is increasing.

Step 1: Find the derivative of the function f(x)

To find the critical points, first, we need to find the derivative of the function \(f(x) = \frac{1}{x-2}\). Let's use the following rule to find the derivative for a function \(\frac{1}{u(x)}\) where \(u(x) = x - 2\): \[ \frac{d}{dx}\left(\frac{1}{u(x)}\right) = -\frac{u'(x)}{[u(x)]^2} \] Applying this rule to our function \(f(x) = \frac{1}{x - 2}\), we get: \[ f'(x) = -\frac{(x-2)'}{[(x-2)]^2} \]

Step 2: Calculate the derivative

Now, we need to calculate the derivative of \(x - 2\): \[ (x-2)' = 1 - 0 = 1 \] Substitute this back to the expression of the derivative, we get: \[ f'(x) = -\frac{1}{[(x-2)]^2} \]

Step 3: Find the critical points

The critical points for the function are the points where the derivative is either equal to zero or does not exist. In case of our function: \[ f'(x) = -\frac{1}{[(x-2)]^2} \] Since there is a rational function with the denominator \((x - 2)^2\), the derivative \(f'(x)\) is not defined at \(x = 2\). Therefore, the critical point is \(x = 2\).

Step 4: Determine if the function is increasing or decreasing

Now we will determine whether the function \(f(x)\) is increasing or decreasing. To do this, we will analyze the intervals around the critical point \(x = 2\). We will test a value on the intervals \((-\infty, 2)\) and \((2, \infty)\) to see if the derivative is positive or negative. 1. For the interval \((-\infty, 2)\), choose a test point \(x = 1\), then: \[ f'(1) = -\frac{1}{[(1-2)]^2} = -1 > 0 \] Since the derivative is negative, the function is decreasing on this interval. 2. For the interval \((2, \infty)\), choose a test point \(x = 3\), then: \[ f'(3) = -\frac{1}{[(3-2)]^2} = -1 > 0 \] Since the derivative is negative, the function is decreasing on this interval as well. In conclusion, the function \(f(x) = \frac{1}{x - 2}\) is decreasing on the intervals \((-\infty, 2)\) and \((2, \infty)\) and there is no interval where the function is increasing.

Key Concepts

Derivative

The concept of a derivative is a fundamental element of calculus. It represents the rate at which a function is changing at any given point. More formally, the derivative of a function at a certain point gives the slope of the tangent line to the function's graph at that point. When you have a function, let's say, of the form \( f(x) \), finding the derivative involves using rules such as the power rule, product rule, quotient rule, or chain rule, depending on the specific characteristics of the function. For a rational function like \( f(x) = \frac{1}{x-2} \), the derivative is found using the quotient rule or the general formula for differentiating a function of the form \(\frac{1}{u(x)}\), which makes it possible to capture the change in function behavior as we alter \(x\).Understanding derivatives is crucial as it forms the basis for many other concepts in calculus, including critical points and determining whether functions are increasing or decreasing.

Critical Points

Critical points are points on a graph where the derivative is either zero or undefined. These points are significant because they can indicate local maximums, minimums, or points of inflection - places where the function could change from increasing to decreasing or vice versa.To find critical points, you first calculate the derivative of the function. In our exercise with \( f(x) = \frac{1}{x-2} \), the derivative was \( f'(x) = -\frac{1}{(x-2)^2} \). Notice that the derivative does not exist at \( x = 2 \) because the function has a vertical asymptote there. This means there is no slope - which is, effectively, a key point of change for the behavior of the function.Once critical points are found, these are utilized to further analyze the function's behavior in terms of increasing and decreasing intervals.

Increasing and Decreasing Functions

A function's increasing and decreasing behavior is determined primarily by the sign of its derivative. When the derivative of a function is positive on an interval, the function is increasing on that interval. Conversely, if the derivative is negative, the function is decreasing.In the given exercise, we determined both intervals were where the function was decreasing since the derivative \( f'(x) = -\frac{1}{(x-2)^2} \) was always negative for any \( x eq 2 \). Here's how we analyzed the behavior:

• In \((-\infty, 2)\), the function decreases because \( f'(x) \) was negative.
• Similarly, in \((2, \infty)\), the function also decreases because \( f'(x) \) remains negative.

A complete understanding of where a function increases or decreases helps in sketching its graph and analyzing its behavior.

Rational Functions

Rational functions are mathematical expressions representing the ratio of two polynomials. They are of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomial expressions.These functions are important in calculus since they often involve asymptotic behavior, which influences how a function behaves near certain values. For instance, \( f(x) = \frac{1}{x-2} \) has a vertical asymptote at \( x = 2 \), meaning the function heads towards infinity or negative infinity as \( x \) approaches \( 2 \). Understanding rational functions requires recognizing where the polynomials in the numerator and the denominator equal zero, as these points can lead to asymptotes or holes in the graph. Recognizing these helps in predicting the behavior of the function graph, which is invaluable for both algebraic manipulation and calculus analysis. Thus, rational functions are key in building a better comprehension of functions and their applications.