Problem 29 - Calculus Early Transcendentals

Question

Use a graphing utility to estimate the absolute maximum and minimum values of \(f \), if any, on the stated interval, and then use calculus methods to find the exact values. \(f(x)=\left(x^{2}-2 x\right)^{2} ;(-\infty,+\infty)\)

Step-by-Step Solution

Answer

Max: 1 at \(x = 1\); Min: 0 at \(x = 0, 2\).

Step 1: Graph the Function

Use a graphing utility to plot the function \( f(x) = (x^2 - 2x)^2 \). Observe the graph to identify the general shape and locate the regions where the maximum and minimum values might occur.

Step 2: Find Critical Points Using Calculus

To find the critical points, first compute the derivative of the function. Set \( f(x) = (x^2 - 2x)^2 \), then let \( u = x^2 - 2x \). We have \( f(x) = u^2 \). The derivative is \( f'(x) = 2u \cdot u'(x) \). Compute \( u'(x) = 2x - 2 \). Thus, \( f'(x) = 2(x^2 - 2x)(2x - 2) \). Solve \( f'(x) = 0 \) to find critical points.

Step 3: Solve the Derivative Equation

Set \( 2(x^2 - 2x)(2x - 2) = 0 \). This occurs when \( x^2 - 2x = 0 \) or \( 2x - 2 = 0 \). Solve these equations: \( x(x-2) = 0 \) gives \( x = 0 \) or \( x = 2 \). The second equation \( 2x - 2 = 0 \) gives \( x = 1 \). So, critical points are \( x = 0, 1, 2 \).

Step 4: Evaluate Function at Critical Points

Calculate \( f(x) \) for each critical point: \( f(0) = (0^2 - 2 \times 0)^2 = 0 \), \( f(1) = (1^2 - 2 \times 1)^2 = 1 \), \( f(2) = (2^2 - 2 \times 2)^2 = 0 \).

Step 5: Determine Absolute Extrema

From the evaluations, the maximum value is \( f(1) = 1 \) and the minimum value is \( f(0) = 0 \) and \( f(2) = 0 \). Thus, the absolute maximum is 1, and the absolute minimum is 0 on the interval \((-\infty, +\infty)\).

Key Concepts

Graphing Utility

The graphing utility is a powerful tool for visually analyzing functions. It allows you to plot a function over a specified interval and examine its behavior. In this exercise, the function given is \( f(x) = (x^2 - 2x)^2 \). Using a graphing utility, you can generate the graph, which helps in identifying potential regions where the function reaches its maximum and minimum values. This visual representation provides an initial rough estimate of where these extrema may occur. Graphing utilities can highlight important features such as intercepts, turning points, and curve shape. By studying these characteristics, students can make educated guesses before confirming results through calculus techniques. Practicing with graphing tools enhances your intuition for functions and prepares you for finding exact solutions.

Derivative

The derivative is a fundamental concept in calculus, representing the rate of change of a function with respect to its variable. In solving this exercise, the derivative of \( f(x) = (x^2 - 2x)^2 \) is crucial. First, we utilize a substitution to simplify the expression. Let \( u = x^2 - 2x \), then \( f(x) = u^2 \), and use the chain rule to differentiate:

• Find \( u'(x) = 2x - 2 \).
• Then, compute \( f'(x) = 2u \cdot u'(x) = 2(x^2 - 2x)(2x - 2) \).

By solving \( f'(x) = 0 \), you discover critical points which are essential to locating extrema. Understanding how to differentiate compound functions and employing differentiation rules are key skills in calculus. This process enables one to understand the mechanics of instantaneous rates of change.

Critical Points

Critical points occur where the derivative of a function equals zero or is undefined. They are pivotal for identifying where a function may reach its local extrema, such as maximum or minimum values. In our example:

• The derivative found was \( f'(x) = 2(x^2 - 2x)(2x - 2) \).
• Set \( f'(x) = 0 \) and solve for \( x \) to obtain critical points: \( x = 0, 1, 2 \).

By plugging these values back into the original function, \( f(x) \), you can determine the output values and analyze the behavior of the function at these points. Finding critical points is a critical skill because it allows you to analyze and interpret the behavior of real-world phenomena modeled by functions, highlighting the turning points where changes in direction occur.

Absolute Maximum and Minimum

The absolute maximum and minimum values of a function indicate the highest and lowest points over a given interval. For the function \( f(x) = (x^2 - 2x)^2 \), identifying these values involves calculating \( f(x) \) at the critical points. From these calculations:

• \( f(0) = 0 \)
• \( f(1) = 1 \)
• \( f(2) = 0 \)

This evaluation shows that the absolute minimum value is 0 at \( x = 0 \) and \( x = 2 \), while the absolute maximum value is 1 at \( x = 1 \).Recognizing absolute extrema is essential in many fields, ranging from engineering to economics, where knowing the highest and lowest possible values can inform critical decision-making processes. Understanding this concept helps predict behavior and make informed judgments based on mathematical models.