Horizontal Tangents & Extrema: A Calculus Deep Dive

by Luna Greco 52 views

Hey guys! Today, we're diving into a super interesting calculus problem: figuring out where the horizontal tangents are and analyzing the extrema of the function f(x) = 2x³ - 27x² + 84x + 10. This might sound a bit intimidating, but trust me, we'll break it down step-by-step so it's totally understandable. We'll be using some key calculus concepts like derivatives and critical points, but don't worry if those terms feel a bit fuzzy right now – we'll make them crystal clear. By the end of this article, you'll not only know how to solve this specific problem but also have a solid understanding of the underlying principles. So, let's jump right in and get our calculus on!

Understanding Horizontal Tangents

Let's start by understanding horizontal tangents. A horizontal tangent occurs at a point on a curve where the slope of the tangent line is zero. Think about it visually: it's where the curve momentarily flattens out before changing direction. In calculus terms, the slope of the tangent line at any point on a curve is given by the derivative of the function at that point. So, to find horizontal tangents, we need to find the points where the derivative of our function, f(x) = 2x³ - 27x² + 84x + 10, is equal to zero. This is a crucial concept, and it forms the foundation for finding extrema (maximum and minimum points) as well. Essentially, we are looking for those 'flat' spots on the graph, where the function isn't increasing or decreasing, just for a fleeting moment. These points are often the key to understanding the overall behavior of the function. To find them, we'll need to roll up our sleeves and get derivatizing! Remember, the derivative tells us the instantaneous rate of change, and where that rate is zero, we've got a horizontal tangent. This is where the magic happens in calculus, where we connect the function's behavior to its rate of change. It's like being a detective, using the clues from the derivative to uncover the secrets of the original function. So, let's put on our detective hats and get started!

Finding the First Derivative

To find the points where the tangent is horizontal, we first need to compute the first derivative of f(x). Remember the power rule? It states that if we have a term like axⁿ, its derivative is naxⁿ⁻¹. Applying this rule to each term in our function, we get:

f'(x) = d/dx (2x³) - d/dx (27x²) + d/dx (84x) + d/dx (10)

Breaking it down term by term:

  • d/dx (2x³) = 3 * 2x² = 6x²
  • d/dx (27x²) = 2 * 27x = 54x
  • d/dx (84x) = 84
  • d/dx (10) = 0 (The derivative of a constant is always zero).

So, putting it all together, the first derivative is:

f'(x) = 6x² - 54x + 84

This quadratic equation, f'(x) = 6x² - 54x + 84, is the key to unlocking the horizontal tangents. It represents the slope of the tangent line at any given point x on the original function. Now, our goal is to find the x-values where this slope is zero. Think of it like finding the roots of the derivative, the points where the rate of change is momentarily paused. These are our potential turning points, the places where the function might switch from increasing to decreasing, or vice versa. Finding the derivative is like taking a snapshot of the function's movement, capturing its speed and direction at every point. And by setting this snapshot to zero, we're pinpointing the moments of stillness, the critical junctures where the function's journey takes a turn. So, with the derivative in hand, we're one step closer to understanding the function's behavior. Now, let's solve for those crucial x-values!

Setting the Derivative to Zero

Now that we have the first derivative, f'(x) = 6x² - 54x + 84, the next step is to set it equal to zero and solve for x. This will give us the x-coordinates of the points where the tangent line is horizontal. So, we have the equation:

6x² - 54x + 84 = 0

To make things easier, let's first simplify the equation by dividing all terms by 6:

x² - 9x + 14 = 0

Now, we have a simpler quadratic equation to solve. We can solve this by factoring. We are looking for two numbers that multiply to 14 and add up to -9. Those numbers are -2 and -7. So, we can factor the equation as follows:

(x - 2)(x - 7) = 0

This gives us two possible solutions for x:

x - 2 = 0 => x = 2 x - 7 = 0 => x = 7

These x-values, x = 2 and x = 7, are our critical points. They are the x-coordinates where the function might have a local maximum or minimum. Think of them as the potential peaks and valleys on the graph of the function. They're like the crucial checkpoints on a roller coaster, the moments where you're at the top of a hill or the bottom of a dip. Setting the derivative to zero is like pausing the roller coaster at these critical points, allowing us to analyze its position and direction. These x-values are the candidates for our extrema, the points where the function changes its behavior. But we're not quite there yet. We need to determine whether these points are actually maxima or minima, or neither. This is where the second derivative test comes into play. So, let's hold onto these x-values and move on to the next step in our calculus adventure!

Finding the Corresponding y-coordinates

We've found the x-coordinates where the tangent lines are horizontal, but to fully describe these points, we need the corresponding y-coordinates. To do this, we plug our x-values (x = 2 and x = 7) back into the original function, f(x) = 2x³ - 27x² + 84x + 10. This will give us the y-values at these critical points, allowing us to pinpoint their exact location on the graph.

For x = 2:

f(2) = 2(2)³ - 27(2)² + 84(2) + 10 f(2) = 2(8) - 27(4) + 168 + 10 f(2) = 16 - 108 + 168 + 10 f(2) = 76 + 16 - 108 f(2) = 86

So, one horizontal tangent point is (2, 86).

For x = 7:

f(7) = 2(7)³ - 27(7)² + 84(7) + 10 f(7) = 2(343) - 27(49) + 588 + 10 f(7) = 686 - 1323 + 588 + 10 f(7) = 1284 - 1323 f(7) = -39

So, the other horizontal tangent point is (7, -39).

Now we have two points, (2, 86) and (7, -39), where the tangent lines are horizontal. These are our potential maxima and minima, the peaks and valleys of our function. But we still don't know for sure whether they are actually maxima or minima, or perhaps neither. Plugging the x-values back into the original function is like taking a ride on the roller coaster, feeling the height of the peaks and the depth of the valleys. These y-values give us the altitude at our critical points, painting a clearer picture of the function's landscape. Now that we have both the x and y coordinates, we can visualize these points on the graph. They're like landmarks in our calculus journey, guiding us towards a complete understanding of the function's behavior. But our journey isn't over yet! We still need to determine the nature of these points – are they maxima, minima, or something else entirely? This is where the second derivative test comes in, our final tool for analyzing the extrema of the function. So, let's move on and put this tool to work!

Analyzing Extrema

Now that we've found the horizontal tangent points, we need to figure out whether they are local maxima, local minima, or neither. These points are known as extrema. A local maximum is a point where the function's value is greater than the values at nearby points, and a local minimum is a point where the function's value is less than the values at nearby points. There are a couple of ways to determine the nature of these points, but we'll use the second derivative test here.

The Second Derivative Test

The second derivative test is a powerful tool for determining the nature of critical points. It uses the sign of the second derivative at a critical point to determine whether the point is a local maximum or a local minimum. Here's the basic idea:

  • If f''(x) > 0 at a critical point, the function is concave up at that point, which means it's a local minimum.
  • If f''(x) < 0 at a critical point, the function is concave down at that point, which means it's a local maximum.
  • If f''(x) = 0 at a critical point, the test is inconclusive, and we need to use another method.

The second derivative is essentially the rate of change of the rate of change. It tells us how the slope of the tangent line is changing. If the second derivative is positive, the slope is increasing, meaning the function is curving upwards like a smile. This indicates a local minimum, a valley in the function's landscape. Conversely, if the second derivative is negative, the slope is decreasing, meaning the function is curving downwards like a frown. This indicates a local maximum, a peak in the function's landscape. The second derivative test is like having a magnifying glass that allows us to zoom in on the critical points and observe the curvature of the function. It's a clever way to distinguish between peaks and valleys, using the function's concavity as our guide. So, to use this test, we first need to find the second derivative of our function. Let's get to it!

Finding the Second Derivative

To use the second derivative test, we need to find the second derivative of our function. The second derivative is simply the derivative of the first derivative. We already found the first derivative to be:

f'(x) = 6x² - 54x + 84

Now, we differentiate this expression to find the second derivative:

f''(x) = d/dx (6x²) - d/dx (54x) + d/dx (84)

Applying the power rule again:

  • d/dx (6x²) = 2 * 6x = 12x
  • d/dx (54x) = 54
  • d/dx (84) = 0

So, the second derivative is:

f''(x) = 12x - 54

This linear equation, f''(x) = 12x - 54, is our key to determining the nature of the critical points. It tells us how the slope of the tangent line is changing at any given point x. The second derivative is like the speedometer of the slope, indicating whether it's increasing (positive) or decreasing (negative). This information is crucial for distinguishing between local maxima and local minima. Think of it as the function's personality, revealing whether it's smiling (concave up, minimum) or frowning (concave down, maximum). Now that we have the second derivative, we can evaluate it at our critical points and see what it tells us about the function's behavior. It's like having a secret code that unlocks the mystery of the extrema. So, let's take this code and apply it to our critical points!

Applying the Second Derivative Test

Now we have the second derivative, f''(x) = 12x - 54, and our critical points, x = 2 and x = 7. We're ready to apply the second derivative test. We'll evaluate the second derivative at each critical point to determine whether it's a local maximum or a local minimum.

For x = 2:

f''(2) = 12(2) - 54 f''(2) = 24 - 54 f''(2) = -30

Since f''(2) is negative (-30 < 0), the function is concave down at x = 2, which means we have a local maximum at the point (2, 86).

For x = 7:

f''(7) = 12(7) - 54 f''(7) = 84 - 54 f''(7) = 30

Since f''(7) is positive (30 > 0), the function is concave up at x = 7, which means we have a local minimum at the point (7, -39).

The second derivative test has revealed the nature of our critical points! At x = 2, the function is frowning, indicating a local maximum. And at x = 7, the function is smiling, indicating a local minimum. This is like reading the function's facial expressions, deciphering its peaks and valleys. The sign of the second derivative is the key to this interpretation, telling us whether the function is curving upwards or downwards. Now, we have a complete picture of the function's behavior at these critical points. We know their location (coordinates) and their nature (maximum or minimum). This is the culmination of our calculus journey, the moment where we fully understand the function's landscape. But remember, this is just a local analysis. There might be other interesting features beyond these critical points. So, always keep exploring and digging deeper into the fascinating world of calculus!

Conclusion

Alright guys, we've successfully found the horizontal tangent points and analyzed the extrema of the function f(x) = 2x³ - 27x² + 84x + 10. We found that there's a local maximum at the point (2, 86) and a local minimum at the point (7, -39). We did this by first finding the first derivative, setting it equal to zero to find the critical points, and then using the second derivative test to determine the nature of these points. This is a classic calculus problem that demonstrates the power of derivatives in analyzing the behavior of functions. By understanding the concepts of horizontal tangents, critical points, and the second derivative test, you can tackle a wide range of similar problems. Remember, calculus is like a powerful toolkit for understanding change and motion. It allows us to analyze the behavior of functions, predict their future, and optimize their performance. The skills we've learned today are fundamental to many areas of science, engineering, and economics. So, keep practicing, keep exploring, and keep pushing the boundaries of your calculus knowledge. The world is full of fascinating functions waiting to be analyzed! And who knows, maybe you'll discover something new and exciting along the way. So, until next time, keep those derivatives flowing and those extrema analyzed! You've got this!