Unveiling Function Behavior: Intervals Of Increase & Decrease

Hey Plastik Magazine readers! Let's dive into some calculus fun today. We're going to explore how to determine where a function increases or decreases. This is super important for understanding the overall shape and behavior of a function. The function we'll be looking at is: f(x) = x^3 - 3x^2 - 24x. Don't worry if this looks intimidating at first; we'll break it down step by step. We'll find the intervals where f is increasing and the intervals where f is decreasing, and we'll express our answers using interval notation. Ready to get started?

(a) Finding Intervals of Increase

First things first, to find where a function is increasing, we need to utilize the power of calculus, specifically the first derivative. The first derivative of a function tells us about the slope of the tangent line at any given point. If the first derivative is positive, the function is increasing; if it's negative, the function is decreasing; and if it's zero, we're likely at a critical point (a potential turning point). So, let's find the first derivative of our function: f(x) = x^3 - 3x^2 - 24x. Using the power rule, the first derivative, denoted as f'(x), becomes: f'(x) = 3x^2 - 6x - 24. See, not so bad, right?

Now, to find the intervals of increase, we need to determine where f'(x) > 0. But before we can do that, we need to find the critical points, i.e., the points where f'(x) = 0 or where f'(x) is undefined (though in our case, f'(x) is a polynomial, so it's defined everywhere). To find the critical points, we set f'(x) = 0 and solve for x: 3x^2 - 6x - 24 = 0. We can simplify this equation by dividing by 3: x^2 - 2x - 8 = 0. Next, let's factor this quadratic equation: (x - 4)(x + 2) = 0. This gives us two critical points: x = 4 and x = -2. These are the x-values where the slope of the tangent line is zero, and they divide the x-axis into intervals where the function's behavior (increasing or decreasing) might change.

We will now need to test the sign of f'(x) in each of these intervals: (-∞, -2), (-2, 4), and (4, ∞). We can do this by picking a test value within each interval and plugging it into f'(x). For the interval (-∞, -2), let's choose x = -3. Then, f'(-3) = 3(-3)^2 - 6(-3) - 24 = 27 + 18 - 24 = 21. Since f'(-3) > 0, the function is increasing on the interval (-∞, -2). For the interval (-2, 4), let's choose x = 0. Then, f'(0) = 3(0)^2 - 6(0) - 24 = -24. Since f'(0) < 0, the function is decreasing on the interval (-2, 4). Finally, for the interval (4, ∞), let's choose x = 5. Then, f'(5) = 3(5)^2 - 6(5) - 24 = 75 - 30 - 24 = 21. Since f'(5) > 0, the function is increasing on the interval (4, ∞). Therefore, the function f(x) is increasing on the intervals (-∞, -2) and (4, ∞). We've successfully identified the increasing intervals! Great job, everyone!

(b) Finding Intervals of Decrease

Now that we've found the intervals where our function f(x) = x^3 - 3x^2 - 24x is increasing, let's switch gears and find the intervals where it's decreasing. Guess what? We've already done most of the heavy lifting in part (a)! Remember how we found the first derivative, f'(x) = 3x^2 - 6x - 24, and determined the critical points x = -2 and x = 4? These points are crucial because they divide the x-axis into intervals where the function's behavior can change from increasing to decreasing, or vice versa. The intervals are (-∞, -2), (-2, 4), and (4, ∞). To find the intervals of decrease, we need to identify where f'(x) < 0. Lucky for us, we've already tested these intervals in part (a)!

As a quick recap, we tested the sign of f'(x) in each interval by picking test values. We found that in the interval (-∞, -2), f'(x) > 0, meaning the function is increasing. In the interval (-2, 4), f'(x) < 0, meaning the function is decreasing. And in the interval (4, ∞), f'(x) > 0, again meaning the function is increasing. Therefore, the function f(x) is decreasing only on the interval (-2, 4). This interval represents the portion of the graph where the function slopes downwards as we move from left to right. It's like a little valley in the graph. We can say that our function f(x) decreases from x = -2 to x = 4. This completes our analysis of the decreasing intervals of the function. We've conquered finding the intervals of increase and decrease! You are all doing fantastic!

Visualization and Understanding

Let's pause for a moment to connect these mathematical results with a visual understanding. Imagine plotting the graph of the function f(x) = x^3 - 3x^2 - 24x. You'll see that the graph rises from negative infinity until it reaches the point x = -2. This confirms our finding that the function is increasing on the interval (-∞, -2). At x = -2, the graph momentarily flattens out (the tangent line has a slope of zero) before starting to descend. As the graph descends, it reaches a minimum point at x = 4. This confirms our finding that the function is decreasing on the interval (-2, 4). After x = 4, the graph begins to climb again, rising to positive infinity. This confirms our finding that the function is increasing on the interval (4, ∞). The critical points, x = -2 and x = 4, are the locations of a local maximum and a local minimum, respectively. These points are where the function changes direction. Understanding these concepts helps you