Extreme Points Of F(x)=-3x^4+6x^2: Second Derivative Test

Hey everyone! Today, we're diving into the fascinating world of calculus to explore how we can pinpoint the extreme points of a function using the second derivative test. Specifically, we'll be looking at the function $f(x) = -3x^4 + 6x^2$. So, buckle up and let's get started!

Understanding the Function: f(x) = -3x^4 + 6x^2

Before we jump into the nitty-gritty of the second derivative test, let's take a moment to understand our function, $f(x) = -3x^4 + 6x^2$. This is a polynomial function of degree 4, which means it has a few key characteristics. First off, the leading coefficient is -3, which is negative. This tells us that the graph of the function will open downwards. Think of it like a frown – the ends of the graph will point down towards negative infinity. Polynomial functions are also smooth and continuous, meaning there are no breaks, jumps, or sharp corners in their graphs. This is super important because the tools of calculus, like derivatives, rely on this smoothness.

The function $f(x) = -3x^4 + 6x^2$ also has some interesting symmetry. Notice that all the powers of x are even (x⁴ and x²). This means that $f(x) = f(-x)$, which tells us that the function is even. Graphically, this means the graph is symmetric about the y-axis. If you were to fold the graph along the y-axis, the two halves would perfectly match up. This symmetry can sometimes help us simplify our analysis, as we might only need to focus on one side of the y-axis.

To find the extreme points, we're looking for the peaks and valleys of the function. These are the points where the function reaches a local maximum or a local minimum. At these points, the slope of the tangent line to the graph is zero. This is a crucial concept because the first derivative of a function gives us the slope of the tangent line at any point. So, to find the extreme points, we'll need to find where the first derivative is equal to zero. However, simply finding where the first derivative is zero isn't enough. We also need to classify these points as either local maxima, local minima, or neither. This is where the second derivative test comes into play. The second derivative tells us about the concavity of the function. If the second derivative is positive at a critical point, the function is concave up (like a cup), and we have a local minimum. If the second derivative is negative, the function is concave down (like a frown), and we have a local maximum. If the second derivative is zero, the test is inconclusive, and we need to use other methods to classify the point.

Step 1: Finding the First Derivative

The first step in using the second derivative test is to find the first derivative of our function. Remember, the first derivative, often denoted as $f'(x)$, gives us the slope of the tangent line to the graph of $f(x)$ at any point. To find the derivative, we'll use the power rule, which states that if $f(x) = ax^n$, then $f'(x) = nax^{n-1}$. Applying this rule to our function, $f(x) = -3x^4 + 6x^2$, we get:

$$f'(x) = -3(4x^3) + 6(2x) = -12x^3 + 12x$$

So, our first derivative is $f'(x) = -12x^3 + 12x$. This new function tells us how the original function $f(x)$ is changing. A positive value of $f'(x)$ means $f(x)$ is increasing, a negative value means $f(x)$ is decreasing, and a zero value indicates a potential extreme point. Speaking of extreme points, the points where $f'(x) = 0$ are called critical points, and they are the candidates for local maxima and minima. To find these critical points, we need to set $f'(x)$ equal to zero and solve for x.

$$0 = -12x^3 + 12x$$

We can factor out a -12x from the equation:

$$0 = -12x(x^2 - 1)$$

And further factor the difference of squares:

$$0 = -12x(x - 1)(x + 1)$$

Now, we can see that the equation is satisfied when x = 0, x = 1, or x = -1. These are our critical points. They are the x-values where the slope of the tangent line is zero, meaning the function might have a peak or a valley at these points. However, we don't know for sure yet whether these points are maxima, minima, or neither. This is where the second derivative comes in. The second derivative will help us classify these critical points by telling us about the concavity of the function at these points. But before we can use the second derivative, we need to find it.

Step 2: Finding the Second Derivative

Now that we've found the first derivative, $f'(x) = -12x^3 + 12x$, the next step is to find the second derivative, denoted as $f''(x)$. The second derivative gives us information about the concavity of the function. Remember, concavity tells us whether the graph of the function is curving upwards (concave up) or downwards (concave down). A positive second derivative indicates concave up, while a negative second derivative indicates concave down. To find the second derivative, we simply take the derivative of the first derivative. Again, we'll use the power rule.

f''(x) = rac{d}{dx}(-12x^3 + 12x) = -12(3x^2) + 12(1) = -36x^2 + 12

So, our second derivative is $f''(x) = -36x^2 + 12$. This function is crucial for the second derivative test. It tells us the rate of change of the slope of the tangent line. A positive $f''(x)$ means the slope is increasing (concave up), and a negative $f''(x)$ means the slope is decreasing (concave down). Now that we have the second derivative, we can use it to classify the critical points we found earlier. We'll plug each critical point into $f''(x)$ and see what sign we get. This will tell us whether the function has a local maximum or a local minimum at that point.

Step 3: Applying the Second Derivative Test

Alright, we've reached the heart of the matter – applying the second derivative test! We have our critical points (x = -1, 0, 1) and our second derivative, $f''(x) = -36x^2 + 12$. Now, we'll plug each critical point into $f''(x)$ to determine the concavity of the function at those points.

Let's start with x = -1:

$$f''(-1) = -36(-1)^2 + 12 = -36(1) + 12 = -36 + 12 = -24$$

Since $f''(-1) = -24$ is negative, the function is concave down at x = -1. This means we have a local maximum at x = -1. To find the y-coordinate of this maximum, we plug x = -1 back into the original function:

$$f(-1) = -3(-1)^4 + 6(-1)^2 = -3(1) + 6(1) = -3 + 6 = 3$$

So, we have a local maximum at the point (-1, 3).

Next, let's check x = 0:

$$f''(0) = -36(0)^2 + 12 = -36(0) + 12 = 0 + 12 = 12$$

Since $f''(0) = 12$ is positive, the function is concave up at x = 0. This means we have a local minimum at x = 0. To find the y-coordinate, we plug x = 0 back into the original function:

$$f(0) = -3(0)^4 + 6(0)^2 = -3(0) + 6(0) = 0$$

So, we have a local minimum at the point (0, 0).

Finally, let's check x = 1:

$$f''(1) = -36(1)^2 + 12 = -36(1) + 12 = -36 + 12 = -24$$

Since $f''(1) = -24$ is negative, the function is concave down at x = 1. This means we have a local maximum at x = 1. To find the y-coordinate, we plug x = 1 back into the original function:

$$f(1) = -3(1)^4 + 6(1)^2 = -3(1) + 6(1) = -3 + 6 = 3$$

So, we have a local maximum at the point (1, 3).

Conclusion: Extreme Points Classified!

Awesome! We've successfully used the second derivative test to find and classify the extreme points of the function $f(x) = -3x^4 + 6x^2$. Let's recap our findings:

• Local Maximum at (-1, 3)
• Local Minimum at (0, 0)
• Local Maximum at (1, 3)

This means our function has two peaks at (-1, 3) and (1, 3), and a valley at (0, 0). These are the points where the function reaches its highest and lowest values in a local neighborhood. By using calculus techniques like the first and second derivatives, we can gain a deep understanding of the behavior of functions and their graphs. Understanding these concepts is super useful in many fields, from physics and engineering to economics and computer science. Keep exploring, keep learning, and keep having fun with math!