Finding Inflection Points Of F(x) = 12x^5 + 60x^4 - 100x^3 + 4
Hey guys! Let's dive into a fun calculus problem today. We're going to tackle finding the inflection points of the function f(x) = 12x⁵ + 60x⁴ - 100x³ + 4. This might sound intimidating, but trust me, we'll break it down step by step so it's super easy to follow. Inflection points are those cool spots on a curve where it changes its concavity – think of it like the point where a smile turns into a frown, or vice versa. To find these points, we'll need to use our calculus skills, specifically derivatives. So, grab your thinking caps, and let's get started!
Understanding Inflection Points
Before we jump into the calculations, let's make sure we're all on the same page about inflection points. Inflection points are points on a curve where the concavity changes. What does concavity mean? Well, imagine you're driving along the curve. If the curve is shaped like a smile (concave up), the car would be in a valley. If the curve is shaped like a frown (concave down), the car would be on a hill. An inflection point is where the road transitions from a valley to a hill, or from a hill to a valley. More formally, a function is concave up if its second derivative is positive, and concave down if its second derivative is negative. Therefore, inflection points occur where the second derivative is either zero or undefined. Finding these points is crucial in understanding the behavior of a function, as they help us visualize its shape and identify key features. In our specific problem, we want to identify the x-values where our given function, f(x) = 12x⁵ + 60x⁴ - 100x³ + 4, changes its concavity. This will involve taking the first and second derivatives, setting the second derivative to zero, and solving for x. We'll also need to check the points where the second derivative is undefined, although in this case, as we're dealing with a polynomial, the second derivative will be defined everywhere. Once we have our potential inflection points, we need to verify that the concavity actually changes at those points. This can be done by checking the sign of the second derivative on either side of each potential inflection point. If the sign changes, then we have confirmed an inflection point. Understanding this process is vital, not only for solving this specific problem but also for analyzing the behavior of functions in general. So let's get into the nitty-gritty and start calculating those derivatives!
Step 1: Find the First Derivative
The first step in finding inflection points is to calculate the first derivative of our function, f(x) = 12x⁵ + 60x⁴ - 100x³ + 4. Remember, the derivative gives us the slope of the tangent line at any point on the curve. We use the power rule here, which states that the derivative of xⁿ is nxⁿ⁻¹. Let's apply this to each term in our function. The derivative of 12x⁵ is 12 * 5x⁴ = 60x⁴. Next, the derivative of 60x⁴ is 60 * 4x³ = 240x³. For the term -100x³, the derivative is -100 * 3x² = -300x². And finally, the derivative of the constant term 4 is simply 0. So, putting it all together, the first derivative, which we'll call f'(x), is: f'(x) = 60x⁴ + 240x³ - 300x². This function represents the slope of the original function at any given x-value. But we need the second derivative to find inflection points, so let's move on to the next step. The first derivative helps us understand where the function is increasing or decreasing, but it doesn't tell us about the concavity, which is what we're after for inflection points. Remember, the first derivative being zero indicates a critical point (a potential maximum or minimum), but an inflection point is a different beast altogether. We need to look at how the slope itself is changing, which is what the second derivative will tell us. This is why the next step, finding the second derivative, is so crucial in our quest to find the inflection points of the function. So, with our first derivative in hand, we're ready to take the next step towards uncovering the inflection points. Let's do it!
Step 2: Calculate the Second Derivative
Okay, we've got the first derivative: f'(x) = 60x⁴ + 240x³ - 300x². Now, we need to find the second derivative, f''(x), which will tell us about the concavity of the function. We'll apply the power rule again to each term in f'(x). The derivative of 60x⁴ is 60 * 4x³ = 240x³. The derivative of 240x³ is 240 * 3x² = 720x². And the derivative of -300x² is -300 * 2x = -600x. So, the second derivative, f''(x), is: f''(x) = 240x³ + 720x² - 600x. This function is key to finding inflection points because it tells us how the rate of change of the slope is changing. Remember, inflection points occur where the concavity changes, and the concavity changes where the second derivative is either zero or undefined. Since we're dealing with a polynomial, our second derivative is defined everywhere, so we just need to find where it equals zero. This means our next step is to set f''(x) equal to zero and solve for x. This will give us the potential x-values where inflection points might occur. However, it's crucial to remember that just because the second derivative is zero doesn't automatically mean we have an inflection point. We still need to verify that the concavity actually changes at those points. But for now, let's focus on finding those potential inflection points by solving the equation f''(x) = 0. We're getting closer to cracking this problem, guys! Hang in there!
Step 3: Set the Second Derivative to Zero and Solve for x
Alright, we've got our second derivative, f''(x) = 240x³ + 720x² - 600x. Now, the crucial step: we need to find where the concavity might change, which means setting the second derivative equal to zero and solving for x: 240x³ + 720x² - 600x = 0. This looks a bit intimidating, but don't worry, we can simplify it. First, notice that all the terms have a common factor of 240x. Let's factor that out: 240x(x² + 3x - 2.5) = 0. This gives us our first potential solution: x = 0. Now we need to solve the quadratic equation: x² + 3x - 2.5 = 0. We can use the quadratic formula for this: x = [-b ± √(b² - 4ac)] / 2a. In our case, a = 1, b = 3, and c = -2.5. Plugging these values in, we get: x = [-3 ± √(3² - 4 * 1 * -2.5)] / 2 * 1 x = [-3 ± √(9 + 10)] / 2 x = [-3 ± √19] / 2. This gives us two more potential solutions: x = (-3 + √19) / 2 ≈ 0.686 and x = (-3 - √19) / 2 ≈ -3.686. So, we have three potential inflection points: x = -3.686, x = 0, and x = 0.686. But remember, these are just potential inflection points. We still need to confirm that the concavity actually changes at these x-values. This means we need to test the intervals around these points to see if the sign of f''(x) changes. We're on the home stretch now! Let's move on to the next step and verify these potential inflection points.
Step 4: Verify Inflection Points by Checking Concavity
Okay, we've identified our potential inflection points: x ≈ -3.686, x = 0, and x ≈ 0.686. Now, the crucial step: we need to verify that these are indeed inflection points by checking if the concavity changes at these x-values. Remember, concavity is determined by the sign of the second derivative, f''(x). To do this, we'll pick test points in the intervals created by our potential inflection points and plug them into f''(x) = 240x³ + 720x² - 600x. Our intervals are: (-∞, -3.686), (-3.686, 0), (0, 0.686), and (0.686, ∞). Let's choose test points within each interval. For (-∞, -3.686), let's pick x = -4. Plugging this into f''(x), we get: f''(-4) = 240(-4)³ + 720(-4)² - 600(-4) = -15360 + 11520 + 2400 = -1440. Since f''(-4) is negative, the function is concave down in this interval. For (-3.686, 0), let's pick x = -1. Plugging this into f''(x), we get: f''(-1) = 240(-1)³ + 720(-1)² - 600(-1) = -240 + 720 + 600 = 1080. Since f''(-1) is positive, the function is concave up in this interval. Notice that the concavity changed at x ≈ -3.686, so this is indeed an inflection point. For (0, 0.686), let's pick x = 0.5. Plugging this into f''(x), we get: f''(0.5) = 240(0.5)³ + 720(0.5)² - 600(0.5) = 30 + 180 - 300 = -90. Since f''(0.5) is negative, the function is concave down in this interval. The concavity changed at x = 0, confirming another inflection point. For (0.686, ∞), let's pick x = 1. Plugging this into f''(x), we get: f''(1) = 240(1)³ + 720(1)² - 600(1) = 240 + 720 - 600 = 360. Since f''(1) is positive, the function is concave up in this interval. The concavity changed at x ≈ 0.686, so this is also an inflection point. We've successfully verified that all three potential points are indeed inflection points! Woohoo!
Step 5: State the Inflection Points
Fantastic! We've done the heavy lifting and found all the inflection points for the function f(x) = 12x⁵ + 60x⁴ - 100x³ + 4. Just to recap, we found the potential inflection points by setting the second derivative equal to zero and solving for x. This gave us x ≈ -3.686, x = 0, and x ≈ 0.686. Then, we verified that these were indeed inflection points by checking the concavity on either side of each point. We found that the concavity changed at each of these x-values, confirming they are inflection points. So, to answer the original question, we can state the inflection points as follows: Reading from left to right: * D ≈ -3.686 * E = 0 * F ≈ 0.686 And there you have it! We've successfully navigated the world of derivatives and concavity to pinpoint the inflection points of our function. Remember, inflection points are crucial for understanding the behavior and shape of a curve. They tell us where the function transitions from curving upwards to curving downwards, or vice versa. This information is invaluable in various fields, from physics to economics, where understanding the behavior of functions is essential. So, pat yourselves on the back, guys! You've conquered a calculus challenge. Keep practicing, and you'll become masters of calculus in no time!
So, there you have it! We've successfully found the inflection points of the function f(x) = 12x⁵ + 60x⁴ - 100x³ + 4. By following the steps of finding the first and second derivatives, setting the second derivative to zero, and verifying the concavity change, we were able to pinpoint the x-values where the function changes its curvature. We identified the inflection points as approximately -3.686, 0, and 0.686. This exercise highlights the power of calculus in analyzing the behavior of functions. Inflection points, in particular, provide valuable insights into the shape and characteristics of a curve. They help us understand where a function transitions from concave up to concave down, or vice versa, which can be crucial in various applications. From optimizing designs in engineering to modeling economic trends, the ability to find and interpret inflection points is a valuable skill. Remember, the key to mastering calculus is practice. The more problems you solve, the more comfortable you'll become with the concepts and techniques. So, keep exploring, keep learning, and keep challenging yourselves. You've got this!
