Understanding Local Maxima and Minima of the Function y x^4 - 4x^3 4x^2

In this comprehensive guide, we will explore the process of finding the local maxima and minima of the function y x^4 - 4x^3 4x^2. We will start from the basics and walk you through the steps to determine the critical points and classify them as local maxima or minima using the first and second derivative tests.

Step 1: Finding the First Derivative

The first step in analyzing the function is to find its first derivative. The first derivative of the function is given by:

y' frac{d}{dx} (x^4 - 4x^3 4x^2) 4x^3 - 12x^2 8x

Step 2: Setting the First Derivative to Zero

To find the critical points, we set the first derivative equal to zero:

4x^3 - 12x^2 8x 0

Factoring out the common term, we get:

4x (x^2 - 3x 2) 0

This equation can be further factored as:

4x (x - 1)(x - 2) 0

Therefore, the critical points are:

x 0 x 1 x 2

Step 3: Using the Second Derivative Test

To determine whether these critical points are local maxima or minima, we use the second derivative test. The second derivative of the function is:

y'' frac{d^2}{dx^2} (x^4 - 4x^3 4x^2) 12x^2 - 24x 8

Next, we evaluate the second derivative at each critical point.

Evaluating at x 0

y''(0) 12(0)^2 - 24(0) 8 8

Since y''(0) > 0, the function has a local minimum at x 0.

Evaluating at x 1

y''(1) 12(1)^2 - 24(1) 8 12 - 24 8 -4

Since y''(1) , the function has a local maximum at x 1.

Evaluating at x 2

y''(2) 12(2)^2 - 24(2) 8 48 - 48 8 8

Since y''(2) > 0, the function has a local minimum at x 2.

Summary of Critical Points

The function y x^4 - 4x^3 4x^2 has:

A local minimum at (0, 0) A local maximum at (1, 1) A local minimum at (2, 0)

Conclusion

We conclude that the function has local maxima and minima at the specified points. Understanding these points helps in visualizing the behavior of the function and finding its critical behavior over different intervals. This approach can be applied to other polynomial functions as well for similar analysis.