Finding Relative Extrema Using The Second Derivative Test

Hey guys! Let's dive into the fascinating world of calculus and tackle a common problem: finding the relative extrema of a function. We'll be using the second derivative test, a powerful tool that can help us identify those peaks and valleys on a graph. Today, we're going to break down how to find the relative maxima and minima of the function $g(x) = 2x^2(1 - x^2)$. If an answer doesn't exist, we'll mark it as DNE (Does Not Exist). Let's get started!

Understanding Relative Extrema

Before we jump into the calculations, it's crucial to understand what relative extrema are. Imagine a roller coaster track. The highest points (crests) are the relative maxima, and the lowest points (troughs) are the relative minima. These points represent the local maximum or minimum values of the function within a specific interval. The second derivative test is a method we use in calculus to find these points. It's based on the idea that the second derivative tells us about the concavity of a function – whether the graph is curving upwards (concave up) or downwards (concave down). To use the second derivative test, we'll need to follow a few key steps, including finding the first and second derivatives of our function, identifying critical points, and evaluating the second derivative at those critical points. By doing this, we'll be able to determine whether each critical point is a relative maximum, a relative minimum, or neither. This test is especially useful because it provides a straightforward way to classify critical points without needing to analyze the intervals around them directly. Relative extrema help us understand the behavior of functions, which is vital in optimization problems across various fields like economics, physics, and engineering. So, let’s get this roller coaster going and see how it works with our function!

Step 1: Find the First Derivative

Our first step in finding relative extrema involves finding the first derivative of the function $g(x) = 2x^2(1 - x^2)$. The first derivative, denoted as $g'(x)$, represents the rate of change of the function. It helps us identify critical points, which are potential locations for relative maxima or minima. To find the derivative, we first need to simplify the function:

$g(x) = 2x^2 - 2x^4$

Now, we can apply the power rule for differentiation, which states that if $f(x) = ax^n$, then $f'(x) = nax^{n-1}$. Applying this rule to our function, we get:

$g'(x) = d/dx (2x^2) - d/dx (2x^4)$ $g'(x) = 4x - 8x^3$

The first derivative, $g'(x) = 4x - 8x^3$, is crucial because it tells us where the function's slope is zero or undefined. These points are the critical points. Understanding the first derivative is like having a map of the function’s hills and valleys. It shows us where the function is increasing (positive slope) and decreasing (negative slope). This knowledge is essential for identifying potential turning points. So, with the first derivative in hand, we're now ready to move on to the next step: finding those all-important critical points. Let's see where our function might be hitting those peaks and valleys!

Step 2: Find the Critical Points

Now that we have the first derivative, $g'(x) = 4x - 8x^3$, our next task is to find the critical points of the function. Critical points are the $x$-values where the first derivative is either equal to zero or undefined. These points are crucial because they are the potential locations of relative maxima and minima. To find the critical points, we set the first derivative equal to zero and solve for $x$:

$4x - 8x^3 = 0$

We can factor out a $4x$ from the equation:

$4x(1 - 2x^2) = 0$

This gives us three possible solutions:

1. $4x = 0$ which implies $x = 0$
2. $1 - 2x^2 = 0$ which implies $2x^2 = 1$, so $x^2 = 1/2$, and thus $x = ±√(1/2) = ±1/√2 = ±√2/2$

So, our critical points are $x = 0$, $x = √2/2$, and $x = -√2/2$. These points are like the coordinates on our roller coaster track where the car might change direction. They are the candidates for our relative maxima and minima. Finding these critical points is a key step in understanding the function’s behavior. With these in hand, we’re ready to take the next step: finding the second derivative, which will help us classify these points as maxima, minima, or neither. Let’s keep this momentum going!

Step 3: Find the Second Derivative

To use the second derivative test, we need to find the second derivative of our function. The second derivative, denoted as $g''(x)$, tells us about the concavity of the function. It indicates whether the graph is curving upwards (concave up) or downwards (concave down). We start with the first derivative we found earlier:

$g'(x) = 4x - 8x^3$

Now, we differentiate this again with respect to $x$. Applying the power rule, we get:

$g''(x) = d/dx (4x) - d/dx (8x^3)$ $g''(x) = 4 - 24x^2$

The second derivative, $g''(x) = 4 - 24x^2$, is our key to unlocking the nature of our critical points. If the second derivative is positive at a critical point, the function is concave up, suggesting a relative minimum. If it's negative, the function is concave down, suggesting a relative maximum. If it’s zero, the test is inconclusive. Think of the second derivative as a curvature detector. It helps us see how the slope is changing, which is crucial for identifying the peaks and valleys of our function. Now that we have this curvature detector, we’re ready to put it to use in the next step: evaluating the second derivative at our critical points. Let's see what our function's curves are telling us!

Step 4: Apply the Second Derivative Test

Now that we have the second derivative, $g''(x) = 4 - 24x^2$, and our critical points, $x = 0$, $x = √2/2$, and $x = -√2/2$, we can apply the second derivative test. This test helps us determine whether each critical point corresponds to a relative maximum, a relative minimum, or neither. We do this by evaluating the second derivative at each critical point:

1. For $x = 0$: $g''(0) = 4 - 24(0)^2 = 4$ Since $g''(0) = 4 > 0$, the function is concave up at $x = 0$, indicating a relative minimum.
2. For $x = √2/2$: $g''(√2/2) = 4 - 24(√2/2)^2 = 4 - 24(1/2) = 4 - 12 = -8$ Since $g''(√2/2) = -8 < 0$, the function is concave down at $x = √2/2$, indicating a relative maximum.
3. For $x = -√2/2$: $g''(-√2/2) = 4 - 24(-√2/2)^2 = 4 - 24(1/2) = 4 - 12 = -8$ Since $g''(-√2/2) = -8 < 0$, the function is concave down at $x = -√2/2$, indicating a relative maximum.

By evaluating the second derivative at each critical point, we’ve classified them as either relative maxima or minima. This is like pinpointing the exact locations of the peaks and valleys on our roller coaster track. The second derivative test gives us a clear way to understand the shape of the function around these points. With these classifications in hand, we’re now ready for the final step: finding the actual values of the relative extrema. Let's see how high and low our function goes!

Step 5: Find the Relative Extrema Values

To complete our analysis, we need to find the actual values of the relative extrema. We do this by plugging our critical points into the original function, $g(x) = 2x^2(1 - x^2)$.

1. For the relative minimum at $x = 0$: $g(0) = 2(0)^2(1 - (0)^2) = 0$ So, the relative minimum value is 0.
2. For the relative maximum at $x = √2/2$: $g(√2/2) = 2(√2/2)^2(1 - (√2/2)^2) = 2(1/2)(1 - 1/2) = 1(1/2) = 1/2$ So, the relative maximum value is 1/2.
3. For the relative maximum at $x = -√2/2$: $g(-√2/2) = 2(-√2/2)^2(1 - (-√2/2)^2) = 2(1/2)(1 - 1/2) = 1(1/2) = 1/2$ So, the relative maximum value is also 1/2.

By plugging our critical points back into the original function, we’ve found the actual heights of our peaks and valleys. These are the relative extrema values. This final step gives us a complete picture of our function’s local behavior. We’ve identified the points where the function reaches its local highs and lows, and we know exactly what those high and low values are. Now, we can confidently state our findings. Let’s summarize our results and provide the final answer to our problem!

Final Answer

After all our hard work, we've successfully found the relative extrema of the function $g(x) = 2x^2(1 - x^2)$. Let's summarize our findings:

• Relative Maxima:
  • At $x = -√2/2$, the relative maximum value is $g(-√2/2) = 1/2$.
  • At $x = √2/2$, the relative maximum value is $g(√2/2) = 1/2$.
• Relative Minimum:
  • At $x = 0$, the relative minimum value is $g(0) = 0$.

So, to answer the original questions:

• Relative maxima: $x = -√2/2$, $x = √2/2$
• Relative minimum: $x = 0$

And there you have it, guys! We've navigated the twists and turns of calculus, using the second derivative test to pinpoint the relative extrema of our function. Remember, this process involves finding the first and second derivatives, identifying critical points, and then using the second derivative test to classify those points. It's a powerful method that helps us understand the behavior of functions, and it's a skill that will come in handy in many areas of math and science. Keep practicing, and you'll become a pro at finding those peaks and valleys! Great job, everyone!