Problem 23 - Thomas Calculus in SI Units

Question

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

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

Step 1: Understand the function and interval

We are given the function \( f(x) = x^2 - 1 \) and the interval \(-1 \leq x \leq 2\). Our objective is to find the absolute maximum and minimum values of this function on this interval.

Step 2: Identify critical points

First, find the derivative of the function: \( f'(x) = 2x \). Set the derivative equal to zero to find critical points: \( 2x = 0 \) leads to \( x = 0 \). Check if the critical point \( x = 0 \) lies within the interval \(-1 \leq x \leq 2\). It does.

Step 3: Evaluate the function at critical points and endpoints

We need to evaluate the function at the critical point and at the endpoints of the interval. Calculate \( f(x) \) at each point:- At \( x = -1 \): \( f(-1) = (-1)^2 - 1 = 0 \)- At \( x = 0 \): \( f(0) = 0^2 - 1 = -1 \)- At \( x = 2 \): \( f(2) = 2^2 - 1 = 3 \)

Step 4: Determine absolute extrema

Compare the values obtained in the previous step. The maximum value is \( f(2) = 3 \) and occurs at \( x = 2 \). The minimum value is \( f(0) = -1 \) and occurs at \( x = 0 \).

Step 5: Graph the function

To graph the function \( f(x) = x^2 - 1 \), note that it is a simple parabola with vertex at \((0, -1)\). It opens upwards given the positive coefficient of \( x^2 \). Plot the points \((2, 3)\), \((0, -1)\), and \((-1, 0)\) on graph paper and connect them smoothly, ensuring the curve is symmetric about the y-axis.

Step 6: Identify points of absolute extrema on the graph

From the graph, mark the points corresponding to the absolute maximum and minimum values. The absolute maximum is at \((2, 3)\) and the absolute minimum is at \((0, -1)\).

Key Concepts

Absolute Extrema

Absolute extrema refers to the absolute highest and lowest values (the maximum and minimum) that a function attains on a given interval. In the context of our example, we're working with the function \( f(x) = x^2 - 1 \) over the interval \(-1 \leq x \leq 2\). This means we want to find the highest and lowest points on the curve within this span of \( x \)-values.
You begin this process by evaluating the function at any critical points as well as at the ends of the interval. In this example, the critical point is at \( x = 0 \), and the endpoints are at \( x = -1 \) and \( x = 2 \). By substituting these \( x \) values back into the function, you get \( f(-1) = 0 \), \( f(0) = -1 \), and \( f(2) = 3 \).

• The absolute minimum value is \( -1 \), which occurs at \( x = 0 \).
• The absolute maximum value is \( 3 \), which occurs at \( x = 2 \).

These are the points where the curve reaches its lowest and highest values on the given interval.

Graphing Functions

Graphing a function helps visualize its behavior across an interval. For \( f(x) = x^2 - 1 \), the function represents a parabola opening upwards, which is a typical shape for a quadratic equation like this.
When graphing such a function, it's useful to identify key characteristics such as:

• The vertex, which is the turning point of the parabola. For \( f(x) \), the vertex is at \( (0, -1) \).
• The direction the parabola opens. Since the coefficient of \( x^2 \) is positive, this parabola opens upwards.

Start plotting by locating the vertex and then adding additional points by choosing values of \( x \), such as the critical point and endpoints identified earlier. For our example:

• Plot points \((-1, 0)\), \((0, -1)\), and \((2, 3)\).

Connect these points to form a smooth, upward-opening parabola. Use symmetry about the y-axis to ensure the curve is accurate.

Critical Points

Critical points occur where the derivative of a function is zero or undefined, and these points represent potential maxima or minima. In the context of the function discussed, the derivative is \( f'(x) = 2x \).
Setting the derivative to zero gives \( 2x = 0 \), resulting in a critical point at \( x = 0 \). This point lies within the given interval \(-1 \leq x \leq 2\), so we must consider it when finding the function's extrema. It reflects a special point where the slope of the tangent to the curve is flat.
Here are the key steps to determine critical points:

• Calculate the derivative of the function.
• Set the derivative equal to zero and solve for \( x \).
• Verify if the solutions are within the interval.

The critical point at \( x = 0 \) was determined in our example and plays an essential role in locating the absolute minimum value of \(-1\) on the graph of the function.