Finding Inflection Points Of F(x) = 12x⁵ + 60x⁴ - 240x³ + 6

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Hey guys! Today, we're diving deep into the fascinating world of calculus to explore the inflection points of a polynomial function. Specifically, we'll be dissecting the function f(x) = 12x⁵ + 60x⁴ - 240x³ + 6. This might seem daunting at first, but trust me, with a little calculus magic, we'll uncover its secrets. Inflection points, those sneaky spots where a curve changes its concavity, are crucial in understanding a function's behavior. They tell us where the graph transitions from curving upwards (concave up) to curving downwards (concave down), or vice versa. Identifying these points helps us sketch the graph accurately and grasp the function's overall shape. We'll embark on a journey through derivatives, setting them to zero, and analyzing the signs to pinpoint these inflection points. Think of it as detective work, but with mathematical tools! Our mission is to find the x-coordinates where the second derivative changes sign. This involves calculating the first and second derivatives, solving for the roots of the second derivative, and then meticulously examining the intervals between these roots. We'll use sign analysis to determine the concavity in each interval, ultimately revealing the inflection points. So, buckle up, grab your calculus gear, and let's embark on this exciting mathematical adventure together! By the end of this article, you'll be a pro at finding inflection points and understanding their significance. We'll break down each step, making it super clear and easy to follow. Let's get started and uncover the hidden inflection points of this polynomial!

Step 1: Finding the First Derivative

The first step in our quest is to find the first derivative of f(x). Remember, the derivative gives us the slope of the tangent line at any point on the curve. This is crucial because the points where the slope changes significantly can lead us to inflection points. To find the first derivative, we'll use the power rule, which states that if f(x) = axⁿ, then f'(x) = nax^(n-1). Applying this rule to each term in our function, f(x) = 12x⁵ + 60x⁴ - 240x³ + 6, we get:

  • The derivative of 12x⁵ is 5 * 12x⁴ = 60x⁴.
  • The derivative of 60x⁴ is 4 * 60x³ = 240x³.
  • The derivative of -240x³ is 3 * -240x² = -720x².
  • The derivative of the constant 6 is 0.

Combining these, the first derivative is f'(x) = 60x⁴ + 240x³ - 720x². This new function, f'(x), tells us the rate of change of f(x). Now, let's move on to the second derivative, which will directly help us identify the inflection points. Understanding the first derivative is like understanding the speed of a car; it tells us how fast the function is changing. But to find the inflection points, we need to understand how the speed itself is changing, which is where the second derivative comes in. So, let's dive into the next step and unveil the second derivative!

Step 2: Unveiling the Second Derivative

Now that we have the first derivative, f'(x) = 60x⁴ + 240x³ - 720x², the next step is to find the second derivative, f''(x). The second derivative tells us about the concavity of the function – whether it's curving upwards (like a smile) or downwards (like a frown). Inflection points occur where this concavity changes. We'll again use the power rule to differentiate f'(x). Let's break it down:

  • The derivative of 60x⁴ is 4 * 60x³ = 240x³.
  • The derivative of 240x³ is 3 * 240x² = 720x².
  • The derivative of -720x² is 2 * -720x = -1440x.

Putting it all together, the second derivative is f''(x) = 240x³ + 720x² - 1440x. This function is key to finding our inflection points. We need to find where f''(x) = 0 or where f''(x) is undefined. In this case, f''(x) is a polynomial, so it's defined everywhere. Thus, we only need to focus on finding the zeros. We'll do this in the next step by setting f''(x) to zero and solving for x. Remember, these x-values are potential inflection points. Understanding the second derivative is like feeling the curves of a road while driving; it tells us how the slope is changing and helps us anticipate turns. So, let's gear up for the next step, where we'll solve for the potential inflection points!

Step 3: Solving for Potential Inflection Points

Our mission now is to find the values of x where the second derivative, f''(x) = 240x³ + 720x² - 1440x, equals zero. These values are our potential inflection points, the spots where the curve might change its concavity. To solve this cubic equation, we'll first factor out the greatest common factor. Notice that all terms are divisible by 240x. Factoring this out, we get:

240x(x² + 3x - 6) = 0

This gives us one solution immediately: x = 0. Now, we need to solve the quadratic equation x² + 3x - 6 = 0. Since it doesn't factor easily, we'll use the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

In our equation, a = 1, b = 3, and c = -6. Plugging these values into the quadratic formula, we get:

x = (-3 ± √(3² - 4 * 1 * -6)) / (2 * 1) x = (-3 ± √(9 + 24)) / 2 x = (-3 ± √33) / 2

So, we have two more potential inflection points:

  • x = (-3 + √33) / 2
  • x = (-3 - √33) / 2

In summary, our potential inflection points are x = 0, x = (-3 + √33) / 2, and x = (-3 - √33) / 2. These are the x-values where the second derivative is zero, but we still need to confirm whether the concavity actually changes at these points. We'll do this in the next step using a sign analysis. Finding these potential inflection points is like locating possible treasure spots on a map; we still need to dig to see if there's gold! So, let's move on to the sign analysis and confirm our findings!

Step 4: Sign Analysis of the Second Derivative

We've identified our potential inflection points: x = 0, x = (-3 + √33) / 2, and x = (-3 - √33) / 2. Now, we need to determine whether the concavity of f(x) actually changes at these points. To do this, we'll perform a sign analysis on the second derivative, f''(x) = 240x³ + 720x² - 1440x. This involves creating a number line and testing intervals between our potential inflection points. First, let's approximate the values of our roots:

  • (-3 - √33) / 2 ≈ (-3 - 5.74) / 2 ≈ -4.37
  • 0
  • (-3 + √33) / 2 ≈ (-3 + 5.74) / 2 ≈ 1.37

Now, we'll create a number line and mark these points. We'll test the intervals: (-∞, -4.37), (-4.37, 0), (0, 1.37), and (1.37, ∞). We'll pick a test value within each interval and plug it into f''(x) to determine the sign:

  1. Interval (-∞, -4.37): Let's test x = -5. f''(-5) = 240(-5)³ + 720(-5)² - 1440(-5) = -30000 + 18000 + 7200 = -4800. The sign is negative, so f(x) is concave down.
  2. Interval (-4.37, 0): Let's test x = -1. f''(-1) = 240(-1)³ + 720(-1)² - 1440(-1) = -240 + 720 + 1440 = 1920. The sign is positive, so f(x) is concave up.
  3. Interval (0, 1.37): Let's test x = 1. f''(1) = 240(1)³ + 720(1)² - 1440(1) = 240 + 720 - 1440 = -480. The sign is negative, so f(x) is concave down.
  4. Interval (1.37, ∞): Let's test x = 2. f''(2) = 240(2)³ + 720(2)² - 1440(2) = 1920 + 2880 - 2880 = 1920. The sign is positive, so f(x) is concave up.

Since the concavity changes at all three points, they are indeed inflection points. So, the inflection points are approximately x = -4.37, x = 0, and x = 1.37. This sign analysis is like checking the weather forecast for our journey; it tells us about the changing conditions ahead and helps us navigate the curves of the function. We've successfully identified the inflection points! Now, let's summarize our findings.

Step 5: Summarizing the Inflection Points

We've journeyed through the world of derivatives and sign analysis, and now we've arrived at our destination: the inflection points of f(x) = 12x⁵ + 60x⁴ - 240x³ + 6. Let's recap our findings:

  1. We found the second derivative: f''(x) = 240x³ + 720x² - 1440x.
  2. We solved for the potential inflection points by setting f''(x) = 0, which gave us:
    • x = 0
    • x = (-3 + √33) / 2
    • x = (-3 - √33) / 2
  3. We performed a sign analysis on f''(x) to confirm that the concavity changes at these points. We found that:
    • At x ≈ -4.37, the concavity changes from down to up.
    • At x = 0, the concavity changes from up to down.
    • At x ≈ 1.37, the concavity changes from down to up.

Therefore, the inflection points of f(x) are located at:

  • x = (-3 - √33) / 2 ≈ -4.37
  • x = 0
  • x = (-3 + √33) / 2 ≈ 1.37

These points mark the spots where the graph of f(x) transitions between curving upwards and curving downwards. Identifying these inflection points gives us a much clearer picture of the function's behavior and shape. We've successfully navigated the calculus landscape and found our treasure! Understanding these inflection points is like having a roadmap of the function; it helps us anticipate its twists and turns. We've now completed our mission, and we can confidently say we've unveiled the inflection points of this polynomial function. Great job, everyone!

Conclusion

In this comprehensive guide, we've successfully navigated the calculus landscape to uncover the inflection points of the function f(x) = 12x⁵ + 60x⁴ - 240x³ + 6. We embarked on a step-by-step journey, starting with finding the first and second derivatives, then solving for potential inflection points, and finally, conducting a sign analysis to confirm our findings. Our hard work paid off, and we identified the inflection points at approximately x = -4.37, x = 0, and x = 1.37. These points are crucial for understanding the function's behavior, as they mark where the concavity changes – from curving upwards to curving downwards, or vice versa. This exploration not only enhances our understanding of calculus but also demonstrates the power of derivatives in analyzing and graphing functions. By mastering these techniques, we can confidently tackle similar problems and gain deeper insights into the world of mathematics. Remember, calculus is not just about formulas and equations; it's about understanding the dynamic relationships between quantities and their rates of change. We hope this guide has illuminated the process of finding inflection points and inspired you to explore further mathematical frontiers. Keep practicing, keep questioning, and keep exploring! The world of calculus is vast and exciting, and there's always more to discover. Until next time, happy calculating!