Unveiling Function Behavior: Rise, Fall, And Extremes

Jan 15, 2026 by Tim Redaksi 54 views

Hey guys! Let's dive into a cool math problem. We've got a function, f(x) = 3x³ - 9x² - 27x + 16, and our mission, should we choose to accept it, is to figure out where this function is going up (increasing), where it's going down (decreasing), find some special spots called stationary points, and then pinpoint the highest and lowest values within a certain range. This is super helpful for understanding how a function behaves! We're going to break down the process step by step, making sure everything is clear and easy to follow. Get ready to explore the exciting world of functions and their ups and downs! It's all about understanding the rate of change of the function and how that translates to its behavior. Let's get started and unravel the mysteries of this cubic function together. We'll be using calculus concepts to get there, but don't worry, I'll make it as simple as possible. This analysis is fundamental for understanding optimization problems, where we are looking for maximum and minimum values, which are key concepts in many fields, including engineering and economics. We need to remember that the derivative is the key to understanding the function. The derivative tells us the slope of the tangent line at any point on the function. When the derivative is positive, the function is increasing; when it's negative, the function is decreasing; and when it's zero, we've found a stationary point! Finding these points helps us to find the range within which the function is increasing or decreasing.

Finding the Function's Derivative

Alright, first things first, we need to find the derivative of our function. The derivative tells us the slope of the function at any given point. To do this, we'll use the power rule, which is a fundamental tool in calculus. For each term in our function f(x) = 3x³ - 9x² - 27x + 16, we apply the power rule: bring down the exponent and multiply it by the coefficient, and then reduce the exponent by one. Let's break it down step by step: The derivative of 3x³ is 9x² (because 3 * 3 = 9 and 3 - 1 = 2). The derivative of -9x² is -18x (because 2 * -9 = -18 and 2 - 1 = 1). The derivative of -27x is -27 (because the derivative of x is 1, so -27 * 1 = -27). The derivative of the constant 16 is 0 (because constants don't change). So, putting it all together, the derivative f'(x) is 9x² - 18x - 27. This is the most crucial step because it sets the stage for everything else we'll do. We'll use this derivative to determine where the function is increasing, decreasing, and where it has stationary points. This understanding forms the backbone of the analysis of this function, giving us a complete view of how it behaves across the given interval. The derivative is more than just a calculation; it provides us with information about the rate of change of the function at any point. Finding the derivative is our key to unlocking the function's secrets. Think of the derivative as the GPS of our function, guiding us through its ups, downs, and plateaus. The ability to find the derivative is a core skill in calculus and serves as the foundation for further explorations, like determining concavity and inflection points. Let's keep our eyes on the prize and compute this derivative correctly, because it's our compass.

Determining Increasing and Decreasing Intervals

Now that we have the derivative f'(x) = 9x² - 18x - 27, we can figure out where the original function f(x) is increasing or decreasing. Remember, if f'(x) > 0, the function is increasing, and if f'(x) < 0, the function is decreasing. To find these intervals, we first need to determine the critical points, which are the points where the derivative is equal to zero or undefined. In this case, f'(x) is a polynomial, so it's defined everywhere. Therefore, we only need to solve for when f'(x) = 0. So, let's solve 9x² - 18x - 27 = 0. First, we can simplify this equation by dividing everything by 9, which gives us x² - 2x - 3 = 0. Next, we can factor this quadratic equation into (x - 3)(x + 1) = 0. This tells us that the critical points are x = 3 and x = -1. These critical points divide the x-axis into intervals. We will use these intervals and test the derivative within each interval. To check the intervals, we pick a test value within each interval and substitute it into f'(x). We can use these points on the x-axis to create intervals: (-∞, -1), (-1, 3), and (3, ∞). For (-∞, -1), let's test x = -2. Then, f'(-2) = 9(-2)² - 18(-2) - 27 = 36 + 36 - 27 = 45. Since f'(-2) > 0, the function is increasing on (-∞, -1). For (-1, 3), let's test x = 0. Then, f'(0) = 9(0)² - 18(0) - 27 = -27. Since f'(0) < 0, the function is decreasing on (-1, 3). For (3, ∞), let's test x = 4. Then, f'(4) = 9(4)² - 18(4) - 27 = 144 - 72 - 27 = 45. Since f'(4) > 0, the function is increasing on (3, ∞). This tells us where the function is going up and where it's going down. Specifically, the function is increasing on the intervals (-∞, -1) and (3, ∞), and it's decreasing on the interval (-1, 3). Now, we're building a clear picture of the function’s behavior. Remember, a positive derivative indicates an increasing function, while a negative derivative indicates a decreasing function. By using the critical points, we identify the turning points of our function, which are crucial for finding the maximum and minimum values. Understanding the intervals of increase and decrease gives us insight into the shape of our curve and its overall trend. We are creating a roadmap of our function's movements on the x-axis. Using the sign of the derivative, we can map out our function's journey, identifying where it climbs, descends, and pauses.

Identifying Stationary Points

Alright, let's zero in on those stationary points. These are the points where the function momentarily stops increasing or decreasing. Mathematically, these occur when the derivative f'(x) = 0. We've already found these points when determining the increasing and decreasing intervals. We've found the critical points to be x = -1 and x = 3. So we have two stationary points. To find the y-coordinate of these points, we plug these x-values back into the original function f(x) = 3x³ - 9x² - 27x + 16. When x = -1, f(-1) = 3(-1)³ - 9(-1)² - 27(-1) + 16 = -3 - 9 + 27 + 16 = 31. So, one stationary point is at (-1, 31). When x = 3, f(3) = 3(3)³ - 9(3)² - 27(3) + 16 = 81 - 81 - 81 + 16 = -65. So, the other stationary point is at (3, -65). Knowing these points is critical because they can represent local maximum or minimum values. We have to be careful, though, because just because a point is stationary doesn’t automatically mean it's a maximum or minimum, but it’s a strong indication. Now we have the coordinates of our stationary points, we can determine the local maximum and minimum by using the first derivative test. We know that x = -1 has a local maximum at (-1, 31) because the function changes from increasing to decreasing. Also, we know that x = 3 has a local minimum at (3, -65) because the function changes from decreasing to increasing. Remember, stationary points are essential because they signify where the function's rate of change is momentarily zero. Finding these points gives us more detail of the function’s shape, which is extremely useful. Identifying stationary points helps us understand how the function changes direction. They are key to finding the turning points of the function, where it can potentially reach its peak or trough. Analyzing these points helps us understand the entire scope of the curve!

Finding Maximum and Minimum Values on the Interval [1, 4]

Now, let's find the maximum and minimum values of the function on the interval [1, 4]. We need to evaluate the function at the critical points within the interval and at the endpoints of the interval. We know that x = 3 is within the interval [1, 4]. We already know f(3) = -65. Now, let's evaluate the function at the endpoints of the interval. When x = 1, f(1) = 3(1)³ - 9(1)² - 27(1) + 16 = 3 - 9 - 27 + 16 = -17. When x = 4, f(4) = 3(4)³ - 9(4)² - 27(4) + 16 = 192 - 144 - 108 + 16 = -44. Now we have three points to evaluate: (3, -65), (1, -17) and (4, -44). Looking at these values, the maximum value on the interval [1, 4] is -17, which occurs at x = 1, and the minimum value on the interval [1, 4] is -65, which occurs at x = 3. Now we have successfully determined the maximum and minimum values of the function within the specified interval. Always remember to check endpoints. They are frequently key to finding the absolute maximum or minimum. Evaluating the function at the endpoints is the final step, ensuring we don't miss any peaks or valleys that might occur at the beginning or end of the interval. By considering these values, we determine the function's absolute highest and lowest points within the specific boundaries.

Summary of Results

Here’s a summary of what we've found:

Increasing Intervals: (-∞, -1) and (3, ∞)
Decreasing Interval: (-1, 3)
Stationary Points: (-1, 31) and (3, -65)
Maximum Value on [1, 4]: -17 at x = 1
Minimum Value on [1, 4]: -65 at x = 3

We have successfully analyzed the function's behavior, determining its increasing and decreasing intervals, identifying the stationary points, and finding the maximum and minimum values within the specified interval. This approach helps you fully understand the function's complete scope. This understanding of a function's behavior is fundamental in calculus, providing insight into its properties and facilitating problem-solving in numerous areas of science and engineering. This methodical approach can be applied to many functions, so keep practicing to master this skill. Congratulations, you've conquered another calculus problem! You can use the knowledge gained from this exercise in many different problems that you may encounter in the future. Keep practicing, and you'll find these types of problems easier with each try! Keep up the great work, and keep exploring the amazing world of mathematics! Keep in mind, this analysis gives us a detailed view of how the function is changing and helps us to solve real-world problems. Great job, guys!