Problem 17 - Thomas Calculus

Question

In Exercises \(15-30\) , 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 max is 3 at \( x = 2 \), min is -1 at \( x = 0 \).

Step 1: Find Critical Points

The first step is to find the critical points by taking the derivative of the function and setting it equal to zero. The derivative of \( f(x) = x^2 - 1 \) is \( f'(x) = 2x \). Setting \( 2x = 0 \), we solve to find the critical point \( x = 0 \).

Step 2: Evaluate Function at Critical Points and Endpoints

Next, we need to evaluate the function at the critical point and at the endpoints of the interval \([-1, 2]\). So, calculate \( f(-1) = (-1)^2 - 1 = 0 \), \( f(0) = 0^2 - 1 = -1 \), and \( f(2) = 2^2 - 1 = 3 \).

Step 3: Determine Absolute Maximum and Minimum

The absolute max and min will occur at the calculated points. From our evaluations, \( f(-1) = 0 \), \( f(0) = -1 \), \( f(2) = 3 \). Thus, the absolute maximum value is 3 at \( x = 2 \) and the absolute minimum value is -1 at \( x = 0 \).

Step 4: Graph the Function and Identify Extrema Points

Finally, graph the function \( f(x) = x^2 - 1 \) over the interval \([-1, 2]\). Plot the points \((-1, 0)\), \((0, -1)\), and \((2, 3)\). The graph is a parabola opening upwards, intersecting the y-axis at \( -1 \) and having horizontal extrema at the points calculated.

Key Concepts

Critical Points

Critical points are key to finding the extremum of a function. These points occur where the derivative equals zero or is undefined. For our function, which is a polynomial function, the derivative is always defined. To find the critical points for the function \( f(x) = x^2 - 1 \), we first calculate the derivative \( f'(x) = 2x \).
Then, we set the derivative equal to zero to find potential extremum points:

• \( 2x = 0 \)

Solving this equation, we find \( x = 0 \) is our critical point. At this point, changes in the behavior of the function could signify a maximum or minimum value, which is what we are seeking to identify.

Derivative

The derivative of a function provides insight into the function's rate of change at any point. It is a fundamental tool in calculus, especially when determining critical points and understanding where these points may occur. For the quadratic function \( f(x) = x^2 - 1 \), the derivative \( f'(x) = 2x \) tells us how the function's value will grow or shrink as \( x \) changes.

• A positive derivative indicates the function is increasing.
• A zero derivative indicates a potential maximum or minimum, known as a critical point.
• A negative derivative indicates the function is decreasing.

By understanding these changes, we gain insights into where extreme values occur in a function.

Function Evaluation

Evaluating a function means calculating the function's value at specific points, often critical points and endpoints when dealing with intervals. In our example, we evaluated \( f(x) = x^2 - 1 \) at:

• The critical point \( x = 0 \): \( f(0) = 0^2 - 1 = -1 \)
• The endpoint \( x = -1 \): \( f(-1) = (-1)^2 - 1 = 0 \)
• The endpoint \( x = 2 \): \( f(2) = 2^2 - 1 = 3 \)

By comparing these values, we identified that the absolute minimum value is \(-1\) at \( x = 0 \), and the absolute maximum value is \( 3 \) at \( x = 2 \). Function evaluation allows us to concretely find these extrema.

Interval

An interval in mathematics refers to a specific range of values that \( x \) can take. For our problem, the interval is \([-1, 2]\). Understanding the interval is important because:

• Endpoints need to be evaluated to determine absolute extrema.
• The behavior of the function within the interval (i.e., increase or decrease) provides insight into where the extrema may lie.

The interval affects both the critical points, which must fall within it, and the behavior of the function as well as its maximum and minimum values. With \([-1, 2]\), both endpoints are integral in our function evaluation process to ensure all potential extrema are accounted for.

Graphing Functions

Graphing the function \( f(x) = x^2 - 1 \) over a specific interval is a visual way to observe where extrema occur. Graphing encompasses both plotting critical points and marking absolute extrema:

• The shape of \( f(x) = x^2 - 1 \) is a parabola opening upwards.
• Through the vertex at \( x = 0 \) and crossings at the endpoints \((-1, 0)\) and \((2, 3)\), we see the relation of values clearly.
• The graph visually confirms that the minimum value is \(-1\) at \((0, -1)\) and maximum is \(3\) at \((2, 3)\).

Graphing functions is critical for seeing the big picture and confirming analytical evaluations visually.