Hey guys! Today, we're diving into a fun little math problem. We're going to explore how to graph the function f(x) = (144 - x2)(-2) and then figure out the area tucked neatly under the curve within the interval [0, 6]. Sounds exciting, right? Let's break it down step by step. Understanding the nuances of graphing functions and calculating areas under curves is super important in calculus and has tons of real-world applications, from physics to economics. So, buckle up, and let's get started!
Understanding the Function
Before we even think about graphing or areas, let's really understand our function, f(x) = (144 - x2)(-2). The first thing you'll notice is that negative exponent. Remember what that means? It tells us that we're actually dealing with a reciprocal. So, we can rewrite our function as:
f(x) = 1 / (144 - x2)2
Now, things start to look a bit clearer. We have a fraction, and the denominator is a squared term. This gives us some immediate clues about the behavior of the function. We need to be mindful of when the denominator might be zero, as that would make our function undefined. To figure this out, let's set the inside of the parenthesis equal to zero:
144 - x^2 = 0
Solving for x, we get:
x^2 = 144 x = ±12
Okay, so we know that our function is undefined at x = 12 and x = -12. These are vertical asymptotes, meaning our graph will approach these lines but never actually touch them. This is crucial information for sketching the graph accurately. Also, notice that we're interested in the interval [0, 6]. Both asymptotes lie outside this interval, which simplifies our task a bit, but we still need to consider how the function behaves as x gets closer to these asymptotes. Within our interval, the function will be continuous and well-behaved, making it easier to analyze. The square in the denominator also ensures that the function will always be positive within its domain, which is another vital piece of information for graphing. By carefully analyzing the function's equation, we can predict its general shape and identify key features, making the graphing process much smoother and more intuitive. This step is often overlooked, but it's the foundation for accurate graphing and area calculations.
Graphing the Function
Alright, now comes the fun part – graphing the function f(x) = 1 / (144 - x2)2 on the interval [0, 6]. We already know some key things: it's always positive (because of the square in the denominator), and it has vertical asymptotes at x = ±12. Since our interval is [0, 6], we only need to focus on the behavior of the graph between these points. Let's think about what happens as x moves from 0 to 6.
At x = 0, we have:
f(0) = 1 / (144 - 02)2 = 1 / 144^2 = 1 / 20736
That's a pretty small number, close to zero. So, our graph starts very close to the x-axis. Now, as x increases, the term (144 - x^2) decreases, which means (144 - x2)2 also decreases. Since this term is in the denominator, the overall value of f(x) will increase. In simpler terms, the graph will start rising as we move to the right.
What happens at x = 6?
f(6) = 1 / (144 - 62)2 = 1 / (144 - 36)^2 = 1 / 108^2 = 1 / 11664
This value is larger than f(0), which confirms that the graph is indeed rising. However, it's still a relatively small number. The graph will curve upwards gradually within our interval. To get a better sense of the curve, we could calculate a few more points. For example, let's try x = 3:
f(3) = 1 / (144 - 32)2 = 1 / (144 - 9)^2 = 1 / 135^2 = 1 / 18225
This point falls between f(0) and f(6), as we'd expect. Now, if you were to plot these points and sketch the curve, you'd see a gentle upward slope, starting very close to the x-axis at x = 0 and gradually rising until x = 6. The graph will be smooth and continuous within our interval, without any sharp corners or breaks. Remember, it won't cross the x-axis because the function is always positive. Using graphing software or even just a rough sketch by hand helps to visualize the behavior of the function, which is super important for understanding the area we're about to calculate.
Finding the Area Under the Curve
Okay, guys, this is where things get a little more intense, but don't worry, we'll tackle it together. We need to find the area under the curve of f(x) = 1 / (144 - x2)2 on the interval [0, 6]. This means we need to calculate the definite integral:
∫[0 to 6] (1 / (144 - x2)2) dx
This integral looks a bit intimidating, right? It's not one we can solve with simple techniques. We're going to need some clever tricks up our sleeves, specifically, trigonometric substitution. The form of the denominator, (144 - x2)2, should give you a hint that a trigonometric substitution might be helpful. We can rewrite 144 as 12^2, so we have the form (a^2 - x^2), where a = 12. This suggests the substitution:
x = 12sin(θ)
Why this substitution? Because it allows us to simplify the expression (144 - x^2) using the trigonometric identity cos^2(θ) = 1 - sin^2(θ). Let's see how it works:
If x = 12sin(θ), then dx = 12cos(θ) dθ. Now we need to rewrite the denominator:
144 - x^2 = 144 - (12sin(θ))^2 = 144 - 144sin^2(θ) = 144(1 - sin^2(θ)) = 144cos^2(θ)
So, (144 - x2)2 = (144cos2(θ))2 = 144^2 cos^4(θ). Our integral now becomes:
∫ (1 / (144^2 cos^4(θ))) * 12cos(θ) dθ = (12 / 144^2) ∫ (1 / cos^3(θ)) dθ = (1 / 1728) ∫ sec^3(θ) dθ
Ah, much better! We've transformed our integral into something we can work with. The integral of sec^3(θ) is a standard integral, but it requires another trick: integration by parts. The integral of sec^3(θ) is 1/2 [sec(θ)tan(θ) + ln|sec(θ) + tan(θ)|]. We also need to adjust our limits of integration. When x = 0, sin(θ) = 0, so θ = 0. When x = 6, sin(θ) = 6/12 = 1/2, so θ = π/6. After evaluating the integral with the limits, we should get the area under the curve.
Solving the Integral and Finding the Area
Okay, let's continue from where we left off. We had transformed our original integral into:
(1 / 1728) ∫[0 to π/6] sec^3(θ) dθ
We know that the integral of sec^3(θ) is given by:
∫ sec^3(θ) dθ = (1/2) [sec(θ)tan(θ) + ln|sec(θ) + tan(θ)|] + C
So, we need to evaluate this from θ = 0 to θ = π/6. Let's plug in the limits. First, let's consider θ = π/6:
sec(π/6) = 1 / cos(π/6) = 1 / (√3/2) = 2/√3 = (2√3)/3 tan(π/6) = sin(π/6) / cos(π/6) = (1/2) / (√3/2) = 1/√3 = √3/3
Now, let's plug in θ = 0:
sec(0) = 1 / cos(0) = 1 / 1 = 1 tan(0) = sin(0) / cos(0) = 0 / 1 = 0
Now, let's substitute these values into our integral result:
(1 / 1728) * (1/2) * [(2√3/3)(√3/3) + ln|(2√3/3) + (√3/3)| - (1 * 0 + ln|1 + 0|)]
Simplify the expression:
(1 / 3456) * [(2/3) + ln|√3| - 0] = (1 / 3456) * [(2/3) + ln(√3)]
We can simplify ln(√3) as (1/2)ln(3), so our final expression for the area is:
(1 / 3456) * [(2/3) + (1/2)ln(3)]
This is the exact value of the area under the curve. If you want a decimal approximation, you can plug this into a calculator. The result will be a small positive number, which makes sense given the graph we sketched earlier. This final calculation brings together all the steps we've taken: understanding the function, graphing it to visualize the area, setting up the definite integral, using trigonometric substitution to simplify the integral, and finally, evaluating the integral to find the area. It's quite a journey, but we made it!
Conclusion
So, guys, we've successfully graphed the function f(x) = (144 - x2)(-2) on the interval [0, 6] and calculated the area under the curve. We started by understanding the function's behavior, identifying its asymptotes, and sketching a graph. Then, we tackled the definite integral, using a clever trigonometric substitution to simplify it. Finally, we evaluated the integral and found the exact area. This whole process really highlights the power of calculus in solving interesting problems. From the initial graph visualization to the intricate integral calculation, each step builds on the previous one, giving us a comprehensive understanding of the function and its properties. Remember, practice makes perfect! The more you work through problems like this, the more comfortable you'll become with these techniques. Keep exploring, keep learning, and most importantly, have fun with math!