Problem 23 - Thomas Calculus

Question

In Exercises \(21-36,\) find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ f(x)=x^{2}-1, \quad-1 \leq x \leq 2 $$

Step-by-Step Solution

Answer

Absolute minimum is -1 at (0, -1), and absolute maximum is 3 at (2, 3).

Step 1: Identify Critical Points

To find the critical points, we first need to find the derivative of the function. The function is given as \[ f(x) = x^2 - 1 \]The derivative of the function is \[ f'(x) = 2x \]We set the derivative equal to zero to find the critical points:\[ 2x = 0 \]Solving for \( x \), we find:\[ x = 0 \]

Step 2: Evaluate Function at Endpoints and Critical Points

Now, let's evaluate the function \( f(x) = x^2 - 1 \) at the critical point \( x = 0 \), and at the endpoints \( x = -1 \) and \( x = 2 \).For \( x = -1 \):\[ f(-1) = (-1)^2 - 1 = 0 \]For \( x = 0 \):\[ f(0) = 0^2 - 1 = -1 \]For \( x = 2 \):\[ f(2) = 2^2 - 1 = 3 \]

Step 3: Determine Absolute Maximum and Minimum

By comparing the values- \( f(-1) = 0 \)- \( f(0) = -1 \)- \( f(2) = 3 \)we see that the absolute maximum value is 3, occurring at \( x = 2 \), and the absolute minimum value is -1, occurring at \( x = 0 \).

Step 4: Graph the Function with Critical Points

To graph the function \( f(x) = x^2 - 1 \) along the interval \( -1 \leq x \leq 2 \):1. The graph is a parabola opening upwards with vertex at \( (0, -1) \), which is the minimum point.2. The graph passes through \( (-1, 0) \) and reaches the maximum at \( (2, 3) \).Plot these points and sketch the parabola considering the given interval.

Key Concepts

Absolute Extrema

When you hear the term **Absolute Extrema**, think of the highest and lowest points on a graph of a function within a specific interval. Absolute extrema include both the maximum and minimum values that a function can reach.
For a function like \( f(x) = x^2 - 1 \), we are tasked with finding these extrema on the interval \(-1 \leq x \leq 2\).
These points are crucial because they inform us of the behavior of the function within the given limits.Understanding where absolute maximum and minimum values occur helps in optimizing scenarios—like finding the best possible or worst possible values of a function for practical applications.
In this exercise, the absolute maximum is achieved at \( x = 2 \) with the value \( f(2) = 3 \), whereas the absolute minimum is at \( x = 0 \) with \( f(0) = -1 \). Remember, to locate these points effectively, compare values at the endpoints of the interval and any critical points within it.

Graphing Parabolas

Graphing parabolas is a fundamental skill when working with quadratic functions like \( f(x) = x^2 - 1 \). A parabola is the U-shaped curve that can open upwards or downwards based on the sign of the leading coefficient.
For this function, the parabola opens upwards because of the positive coefficient in front of \( x^2 \). The vertex of this parabola, which is its turning point, is at \( (0, -1) \).
When graphing the function, consider:

• The vertex: \( (0, -1) \) marks the lowest point on the graph for this interval.
• The axis of symmetry, a vertical line through the vertex, which is \( x = 0 \) in this case.
• Additional points like \( (-1, 0) \) and \( (2, 3) \) that help in plotting the curve.

To sketch the graph on the interval \(-1 \leq x \leq 2\), plot these points and draw a smooth curve through them, ensuring the curve reflects the upward-opening nature of the parabola.

Critical Points

**Critical Points** are places on a graph where the derivative is zero or undefined. They are significant as they can signal a change in direction of the graph — that is, locations where a function may have a local maximum or minimum.
To find these points for a function such as \( f(x) = x^2 - 1 \):

• First, compute the derivative: \( f'(x) = 2x \).
• Set the derivative equal to zero: \( 2x = 0 \) leads us to the critical point at \( x = 0 \).

This step is integral to identifying potential locations for absolute or relative extrema on the function.
Critical points, such as that found here, indicate where the function's slope is zero—demonstrating a momentary halt in direction change.

Derivative Analysis

**Derivative Analysis** is all about understanding how a function's rate of change behaves. By finding the derivative, we gain insights about the graph's slope and how it transitions through different intervals.
In our function \( f(x) = x^2 - 1 \),

• The derivative \( f'(x) = 2x \) informs us that the slope of the tangent to the curve changes linearly with \( x \).
• A positive slope for \( x > 0 \) and a negative slope for \( x < 0 \) implies that the function is decreasing to the left of \( x = 0 \) and increasing to the right.

Derivative analysis is key for identifying not just critical points, but also for understanding the monotonous behavior of sections between them. This concept is fundamental in making decisions about where functions maximize or minimize. Paying attention to how derivatives change also helps in predicting trends, ensuring accurate function modeling.