Problem 42 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ k(x)=1 /\left(x^{2}-4\right) $$

Step-by-Step Solution

Answer

Decreasing on \((-\infty, -2)\) and \((0, 2)\); increasing on \((-2, 0)\) and \((2, \infty)\).

Step 1: Understand the Function

The function given is \( k(x) = \frac{1}{x^2 - 4} \). This is a rational function where the denominator can equal zero, leading to undefined values at those points. The function will be undefined at any value of \( x \) that makes \( x^2 - 4 = 0 \).

Step 2: Find Critical Points

The critical points occur where the derivative \( k'(x) \) is zero or undefined. First, we find \( k'(x) \). Let \( f(x) = x^2 - 4 \), then \( k(x) = f(x)^{-1} \). The derivative of \( k(x) \) is \( k'(x) = -\frac{f'(x)}{(f(x))^2}\), where \( f'(x) = 2x \). Thus, \( k'(x) = -\frac{2x}{(x^2 - 4)^2} \). Set \( k'(x) = 0 \) to find critical points: \(-\frac{2x}{(x^2 - 4)^2} = 0\), which gives \( x = 0 \) as a critical point.

Step 3: Determine Undefined Points

Set the denominator equal to zero to find where the function is undefined: \( x^2 - 4 = 0 \). Solving this gives \( x = \pm 2 \). The function is thus undefined at \( x = -2 \) and \( x = 2 \).

Step 4: Create Test Intervals

With the critical point \( x = 0 \) and the undefined points, break the real number line into intervals: \((-\infty, -2)\), \((-2, 0)\), \((0, 2)\), and \((2, \infty)\).

Step 5: Analyze the Sign of the Derivative

Select test points from each interval to determine the sign of \( k'(x) \). - For \((-\infty, -2)\), choose \( x = -3 \), resulting in \( k'(-3) = -\frac{2(-3)}{((-3)^2 - 4)^2} < 0 \). The function is decreasing.- For \((-2, 0)\), choose \( x = -1 \), resulting in \( k'(-1) = -\frac{2(-1)}{((-1)^2 - 4)^2} > 0 \). The function is increasing.- For \((0, 2)\), choose \( x = 1 \), resulting in \( k'(1) = -\frac{2(1)}{(1^2 - 4)^2} < 0 \). The function is decreasing.- For \((2, \infty)\), choose \( x = 3 \), resulting in \( k'(3) = -\frac{2(3)}{(3^2 - 4)^2} > 0 \). The function is increasing.

Step 6: Conclusion of Intervals

From the analysis:- The function is decreasing on \((-\infty, -2)\) and \((0, 2)\).- The function is increasing on \((-2, 0)\) and \((2, \infty)\).

Key Concepts

Critical Points

Critical points in calculus are essentially the values of a variable such as \( x \) where the derivative of a function equals zero or becomes undefined. These points are crucial as they signal where the function could potentially change from increasing to decreasing, or vice versa.

For the given rational function \( k(x) = \frac{1}{x^2 - 4} \), we identified the critical points by deriving \( k'(x) \). The derivative \( k'(x) = -\frac{2x}{(x^2 - 4)^2} \) becomes zero when \( x = 0 \).

Additionally, the function itself is undefined where \( x^2 - 4 = 0 \), i.e.,, at \( x = \pm 2 \). Thus, the critical point from the derivative is \( x = 0 \), while \( x = -2 \) and \( x = 2 \) are points of undefined behavior.

Rational Functions

Rational functions are fractions where both the numerator and denominator are polynomials. They are a common subject in calculus due to their unique properties.

In \( k(x) = \frac{1}{x^2 - 4} \), it's clear that this function will have specific values of \( x \) that make it undefined because division by zero is not possible. Here, \( x^2 - 4 = 0 \) at \( x = \pm 2 \), meaning the function cannot be evaluated at these points.

Behavior around these points is particularly interesting. Rational functions often exhibit vertical asymptotes or holes at points of undefined behavior, significantly affecting the graph's shape and the function's increase or decrease pattern around these areas.

Increasing and Decreasing Intervals

The intervals on which a function is increasing or decreasing give us a great insight into its behavior. We determine these intervals by analyzing the sign of the derivative. If \( k'(x) > 0 \), the function is increasing in that interval. Conversely, if \( k'(x) < 0 \), the function is decreasing.

For \( k(x) = \frac{1}{x^2 - 4} \), we divided the number line into intervals based on the critical point \( x = 0 \) and undefined points \( x = -2 \) and \( x = 2 \). The intervals are:

• Decreasing on \(( -\infty, -2 )\) and \(( 0, 2 )\)
• Increasing on \(( -2, 0 )\) and \(( 2, \infty )\)

We tested the derivative \( k'(x) \) on points within these intervals to determine its sign. This method aids in mapping out how the function behaves across different segments, crucial for understanding the overall behavior of the rational function.