Problem 21 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f \), if any, on the given interval, and state where those values occur. \(f(x)=x^{2}-x-2 ;(-\infty,+\infty)\)

Step-by-Step Solution

Answer

Absolute minimum is \(-\frac{9}{4}\) at \(x=\frac{1}{2}\); no absolute maximum.

Step 1: Understanding the Problem

We are tasked with finding the absolute maximum and minimum values of the function \( f(x) = x^2 - x - 2 \) on the interval \( (-\infty, +\infty) \). This means we need to determine where the function reaches its highest and lowest points.

Step 2: Find the Critical Points

To locate critical points, we first need to find the derivative of the function \( f(x) = x^2 - x - 2 \). The derivative is \( f'(x) = 2x - 1 \). Set this derivative equal to zero and solve for \( x \):\[ 2x - 1 = 0 \]\[ x = \frac{1}{2} \]. The critical point is \( x = \frac{1}{2} \).

Step 3: Evaluate the Function at Critical Points

Substitute the critical point back into the function to find the value: \[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 2 = \frac{1}{4} - \frac{1}{2} - 2 \]. After simplifying, we get \( f\left(\frac{1}{2}\right) = -\frac{9}{4} \). This is the minimum value of the function.

Step 4: Determine Behavior as \( x \to \pm\infty \)

Since the coefficient of \(x^2\) is positive, the function \( f(x) = x^2 - x - 2 \) opens upwards. As \( x \to +\infty \) and \( x \to -\infty \), \( f(x) \to +\infty \). Therefore, the function does not have an absolute maximum.

Key Concepts

Critical Points

Critical points are essential in understanding the behavior of a function. To identify these points, you first need to find the derivative of the function and determine where it's equal to zero or undefined. Critical points can potentially be locations of maximum or minimum values of the function. They occur at values of \( x \) where the derivative \( f'(x) \) becomes zero or does not exist.

• Calculate the derivative of the function.
• Solve \( f'(x) = 0 \) to find the values of \( x \).

If you encounter any values for which the derivative does not exist, these also count as critical points. Each critical point is a candidate for being a local extremum, and further evaluation is necessary to classify them effectively.

Derivative

A derivative represents the slope of the tangent line to the curve of the function at any given point. This mathematical tool allows us to analyze and interpret the rate at which a function changes. For a given function \( f(x) \), its derivative, often expressed as \( f'(x) \) or \( \frac{df}{dx} \), provides critical insights:

• If \( f'(x) > 0 \), the function is increasing at \( x \).
• If \( f'(x) < 0 \), the function is decreasing at \( x \).
• If \( f'(x) = 0 \), it may suggest a local minimum, maximum, or a point of inflection.

When finding critical points, solving for \( f'(x) = 0 \) is crucial, as it helps locate points that warrant further examination for maxima or minima.

Extrema

Extrema refer to the highest or lowest points of a function within a given interval, known as the maximum or minimum values. They are categorized into relative (local) and absolute (global) extrema. A relative extremum is a high or low point in a specific region of the function, while an absolute extremum is the highest or lowest value among the entire range.

• Local extrema occur at critical points where the derivative changes sign.
• Absolute extrema are determined by evaluating the function at critical points and endpoints, if any exist within the interval.

In our exercise, the absolute minimum was found at \( x = \frac{1}{2} \), with \( f\left(\frac{1}{2}\right) = -\frac{9}{4} \). This was determined by calculating the derivative and checking the sign behavior around the critical point. As for the absolute maximum, the function open upwards with \( f(x) \to +\infty \) as \( x \to \pm \infty \), indicating no true absolute maximum exists.