Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find, if possible, the (global) maximum and minimum values of the given function on the indicated interval. $$ H(x)=\left|x^{2}-1\right| \text { on }[-2,2] $$

Step-by-Step Solution

Answer

The maximum value is 3 at \( x = -2 \) and \( x = 2 \); the minimum value is 0 at \( x = -1 \) and \( x = 1 \).

Step 1: Understand the Function

The function given is \( H(x) = |x^2 - 1| \). This function takes any input \( x \), squares it, subtracts one from it, and then takes the absolute value of the result. Our goal is to find the maximum and minimum values of this function on the interval \([-2, 2]\).

Step 2: Analyze Critical Points and Endpoints

In analyzing for maximum and minimum values of a function, we consider the critical points and the values at the endpoints of the interval. Critical points occur where the derivative \( H'(x) \) is zero or undefined. We must differentiate \( H(x) \) and analyze the values at \( x = -2, 0.5, 2 \).

Step 3: Differentiate and Find Critical Points

The first derivative is found by considering that the function \( H(x) = |x^2 - 1| \) changes nature based on whether \( x^2 - 1 \) is positive or negative. For \( x^2 > 1 \), \( H(x) = x^2 - 1 \) and for \( x^2 < 1 \), \( H(x) = 1 - x^2 \). Find where \( x = 0 \) or \( x = 1, -1 \).

Step 4: Evaluate the Function at the Critical Points and Endpoints

Evaluate \( H(x) \) at all critical points and endpoints within the interval. We calculate: \( H(-2) = 3, H(-1) = 0, H(0) = 1, H(1) = 0, H(2) = 3 \).

Step 5: Identify the Maximum and Minimum Values

From our evaluation, the maximum value is \( 3 \), occurring at \( x = -2 \) and \( x = 2 \). The minimum value is \( 0 \), occurring at \( x = -1 \) and \( x = 1 \).

Key Concepts

Absolute Value Function

The absolute value function is an important concept in calculus that plays a role in understanding how to determine the magnitude of a number. It is expressed as \(|x|\), which is defined as:\[\begin{cases} x, & \text{if } x \geq 0 \ -x, & \text{if } x < 0 \end{cases}\] The purpose of this function is to provide the non-negative value of \(x\), essentially stripping any negative sign. When dealing with expressions inside the absolute value, such as \(x^2-1\), it is crucial to consider two cases:

• When \(x^2-1\) is positive: simply use the expression as it is, becoming \(x^2-1\).
• When \(x^2-1\) is negative: use the negative of the expression, \(1-x^2\).

This approach allows proper analysis and differentiation by splitting the function into piecewise components. For the range \([-2,2]\), checking inside that range helps determine where the expression changes from one piece to another, guiding the search for critical points.

Critical Points Analysis

Critical points are a vital concept in calculus for determining the local behavior of a function, which allows us to locate points of interest like peaks and troughs. To find critical points, we look for places where the derivative of the function is either zero or undefined.For the function \(H(x) = |x^2 - 1|\), the procedure begins with understanding when \(x^2 - 1\) crosses zero. By setting \(x^2 - 1 = 0\), we solve for \(x\), giving \(x = \pm 1\). These are the points where the nature of the function changes.We split the domain based on these points and examine the derivative. Step-by-step, the analysis follows:

• For \(x < -1\) or \(x > 1\): treat as \(x^2 - 1\), leading to the derivative \(2x\).
• For \(-1 < x < 1\): treat as \(1 - x^2\), with a derivative \(-2x\).

By testing within these ranges and at the critical points, along with endpoints, the behavior and features of \(H(x)\) become clear, identifying where potential maxima or minima occur.

Global Maximum and Minimum

Finding the global maximum and minimum involves comparing function values at critical points and endpoints. This tells us where the highest and lowest function values within a given interval appear.For \(H(x) = |x^2 - 1|\) on the interval \([-2,2]\), we look at:

• Endpoints: \(H(-2) = 3\) and \(H(2) = 3\).
• Critical points: \(H(-1) = 0\), \(H(1) = 0\), and \(H(0) = 1\).

Evaluating these, we find the highest value of \(H(x)\) occurs at the endpoints, with \(3\) being the global maximum. The lowest value, \(0\), appears at the critical points \(x = -1\) and \(x = 1\), forming the global minimum.In conclusion, identifying global extrema is about checking the overall behavior across the domain. Comparing these values provides a complete picture of the function's range within the set bounds, making it a critical step in solving optimization problems in calculus.