Problem 16 - College Algebra and Calculus: An Applied Approach

Question

Describe the increasing and decreasing behavior of the function. Find the point or points where the behavior of the function changes. \(f(x)=\sqrt{x^{2}-4}\)

Step-by-Step Solution

Answer

The function \(f(x)=\sqrt{x^{2}-4}\) increases on the intervals (-infinity, -2) and (2, infinity) and decreases on the interval (-2, 2). It changes its behavior at points x = -2 (from increasing to decreasing) and x = 2 (from decreasing to increasing).

Step 1: Differentiate the Function

The first step to analyzing a function's behavior is to find the derivative. Here, we have \(f(x)=\sqrt{x^{2}-4}\) which can also be written as \(f(x)=(x^{2}-4)^{1/2}\). Using the chain rule to differentiate it, the derivative \(f'(x)\) becomes \(\frac{1}{2}(x^{2}-4)^{-1/2}*2x\). This simplifies to \(f'(x)=\frac{x}{\sqrt{x^{2}-4}}\).

Step 2: Find the Critical Points

Critical points exist where the derivative is zero or undefined. Here, the derivative \(f'(x)\) will be undefined when the denominator of the fraction is zero i.e. \(x^{2}-4=0\). Solving for x gives x = 2 and x = -2. The derivative is not zero for any x, hence we only have two critical points: x = 2 and x = -2.

Step 3: Analyze the Increasing and Decreasing Behavior

Next, we check the sign of the derivative \(f'(x)\) at points before, between and after the critical points (x = 2 and x = -2). If \(f’(x) > 0\), then f(x) is increasing. If \(f’(x) < 0\), then f(x) is decreasing. Setting up the intervals: (-infinity, -2), (-2, 2), and (2, infinity), choosing any number from these intervals and putting it in \(f'(x)\), we observe that \(f’(x) > 0\) for (-infinity, -2) and (2, infinity) while \(f’(x) < 0\) for (-2, 2). Thus, the function is decreasing on the interval (-2, 2) and increasing on the intervals (-infinity, -2) and (2, infinity).

Step 4: Identify the Points of Behavior Change

As observed in step 3, the function changes its behavior from increasing to decreasing at x = -2 and from decreasing to increasing at x = 2. These are the points where the behavior of the function changes.

Key Concepts

Critical Points in Calculus

In calculus, critical points are essential in understanding the behavior of functions. They are the x-values where the function's derivative is either zero or does not exist. The importance of critical points lies in their ability to indicate potential local maxima, local minima, or points of inflection—where the curve changes concavity.

To find critical points, we take the function's derivative and solve for when it equals zero or when it is undefined. For example, given the function \(f(x)=\sqrt{x^{2}-4}\), we first express it in a form suitable for differentiation, \(f(x)=(x^{2}-4)^{1/2}\). After applying the chain rule for differentiation, we find \(f'(x)=\frac{x}{\sqrt{x^{2}-4}}\), and then determine the values of x that make the derivative undefined. Here, the critical points occur at x = -2 and x = 2 because the square root in the denominator becomes zero, which is undefined. Critical points are indicative of where the function may increase or decrease, but to conclude definitively, we need to analyze the intervals around these points.

Increasing and Decreasing Intervals

The intervals on which a function is increasing or decreasing can reveal much about its overall shape and behavior. After we've identified the function's critical points, the next step is to divide the domain of the function into intervals separated by these critical points and test the sign of the derivative within each interval.

By plugging in representative values from each interval into the derivative, we can determine if the function is increasing (if \(f'(x) > 0\)) or decreasing (if \(f'(x) < 0\)). For the function \(f(x)=\sqrt{x^{2}-4}\), we test points in the intervals (-∞, -2), (-2, 2), and (2, ∞). The derivative's sign suggests that f(x) is increasing on the intervals (-∞, -2) and (2, ∞), while it is decreasing on (-2, 2). These insights are crucial for sketching graphs, predicting function behavior, and solving optimization problems.

Chain Rule Differentiation

The chain rule is a pivotal concept in calculus, especially when dealing with composite functions. It allows us to differentiate a function of a function, effectively peeling back layers of complexity one at a time.

Consider the function \(f(x)=\sqrt{x^{2}-4}\), or re-written, \(f(x)=(x^{2}-4)^{1/2}\). Here, we view \(x^{2}-4\) as the inner function and the square root as the outer function. The chain rule informs us to first differentiate the outer function, which gives us \((1/2)(x^{2}-4)^{-1/2}\), and then multiply by the derivative of the inner function, which is \(2x\). This results in the derivative \(f'(x)=\frac{x}{\sqrt{x^{2}-4}}\).

Without the chain rule, analyzing the behavior of composite functions would be a complex task. By using it, we can clearly find critical points and determine where the function increases or decreases, simplifying the study of the function’s behavior significantly.