Optimizing a Quadratic Function: Finding Extrema in a Given Interval

This article will guide you through the process of identifying the extrema of a quadratic function, (f(x) x^2 x^2), defined on the interval [1, 3]. We will explore various techniques and mathematical operations to accurately determine whether these extrema are local or global.

Introduction

The function (f(x) x^2 x^2) appears unusual due to the duplication of the term, but for the sake of this article, let's focus on a more standard form: (f(x) x^2 x^2 2x^2). We will use a transformed form of this function to simplify the process. By defining a new function, (g(x) x^2 - x - 2), we can identify and classify the extrema more efficiently.

Step 1: Transform the Function for Analysis

The first step is to transform the function into a more manageable form. By subtracting and adding terms, we get (g(x) x^2 - x - 2). This step is crucial as it allows us to use factoring and other algebraic techniques to find the roots of (g(x)).

Step 2: Find and Classify the Roots

The second step involves finding the roots of the transformed function, (g(x)). Factoring (g(x)) gives us:

(g(x) x^2 - x - 2 (x - 2)(x 1))

The roots are (x 2) and (x -1). We are interested in the roots that lie within the interval [1, 3]. Therefore, (x 2) is our only relevant root.

Step 3: Determine the Interval and Nature of the Extrema

Now that we have the root (x 2), we can divide the interval [1, 3] into two parts: [1, 2] and [2, 3]. The root (x 2) acts as a critical point within this interval. By evaluating (g(x)) at the endpoints and at the root, we can determine the nature of the extrema.

Evaluating (g(x)) at the endpoints and the root gives us:

(g(1) 1^2 - 1 - 2 -2)

(g(2) 2^2 - 2 - 2 0)

(g(3) 3^2 - 3 - 2 4)

From these evaluations, we can conclude that (x 2) is a local minimum because the value at (x 2) is lower than the values at the endpoints (x 1) and (x 3).

Step 4: Verify the Nature of Critical Points

After identifying the suspicious points, we need to verify whether these points are maxima or minima. We do this by taking the first and second derivatives of (g(x)).

(g'(x) 2x - 1)

(g''(x) 2)

By setting (g'(x) 0), we find another potential critical point:

(2x - 1 0 Rightarrow x frac{1}{2})

However, (x frac{1}{2}) is not within the interval [1, 3], so we do not consider it.

The second derivative test confirms that (g''(x) 2), which is positive, indicating that (g(x)) has a local minimum at (x 2). Since this is the only critical point in the interval, it is also the global minimum.

Conclusion

By transforming the original function and using algebraic techniques, we can efficiently find and classify the extrema of a quadratic function within a specified interval. In this case, the function (f(x) x^2 x^2) (simplified to (f(x) 2x^2)) has a global minimum at (x 2) within the interval [1, 3].

Additional Resources

For readers interested in further exploration, here are some additional resources:

Khan Academy - Exploring Extrema of a Function Math Is Fun - Extreme Value of a Function Test-Guide - Optimizing Functions with Calculus