Calculate F'(π) + F(π) + 2: Math Problem Solution

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Hey guys, let's dive into this calculus problem! We're tasked with figuring out the value of f'(π) + f(π) + 2, and we're given the function f(x) = x sin(2x). It might seem a bit daunting at first, but trust me, it's totally manageable if we break it down step by step. We'll need to find the derivative of the function, evaluate the function and its derivative at x = π, and then plug those values into the final expression. Let's get started!

Understanding the Problem: The Core Concepts

Alright, before we jump into the nitty-gritty, let's make sure we're all on the same page with the core concepts. This problem beautifully blends differentiation and evaluation. We need to find the derivative of a function (which represents its instantaneous rate of change), evaluate both the original function and its derivative at a specific point (π in this case), and finally, combine those results with a simple addition. This problem is a classic example of how calculus ties together different ideas. Make sure you are familiar with the product rule of derivatives because we are going to use it.

First, we've got our function f(x) = x sin(2x). The key here is recognizing that it's a product of two functions: x and sin(2x). That means we'll need to use the product rule to find the derivative f'(x). Remember the product rule? It says that if we have a function h(x) = u(x)v(x), then its derivative is h'(x) = u'(x)v(x) + u(x)v'(x). We can break it down, no sweat. We'll then need to evaluate the original function and its derivative at x = π. This means we'll substitute π in for every 'x' we see in the equations we come up with. Easy peasy!

This kind of problem is fundamental in calculus, and mastering it helps a lot in your understanding of the subject. You'll see this kind of stuff popping up in physics, engineering, and pretty much anywhere you need to model change. So, by nailing this problem, you're not just solving one math problem – you're building a foundation for lots more to come. You are not only going to be able to find the derivative of more complex functions but it will also help with the evaluation of these functions. So, let's keep going and ace it!

Finding the Derivative: Applying the Product Rule

Now, let's find the derivative, f'(x). Since f(x) = x sin(2x), we can consider u(x) = x and v(x) = sin(2x). This is where the product rule shines! The derivative of u(x) (which is just x) is u'(x) = 1. For v(x) = sin(2x), we'll need the chain rule. The derivative of sin(2x) is cos(2x) multiplied by the derivative of the inside function, which is 2. So, v'(x) = 2cos(2x). Then, using the product rule, the derivative is: f'(x) = u'(x)v(x) + u(x)v'(x). Substituting the values: f'(x) = 1 * sin(2x) + x * 2cos(2x), which simplifies to f'(x) = sin(2x) + 2x cos(2x). It's really not too scary, right? You just have to follow the formulas. It's like a recipe; you gotta have all the ingredients in the right places.

So, we've successfully found the derivative of our function, and we've done it correctly. Now we have two key equations to work with. We will use them to help solve our original equation. Don't worry though, because this is the hardest part. The rest is just plugging in numbers. Always remember, the more practice you put in the better you will get. Understanding this process is really beneficial and we are almost done!

Evaluating at x = π: Plugging in the Values

Now for the fun part: let's evaluate f(x) and f'(x) at x = π. For f(x) = x sin(2x), we substitute x with π: f(π) = π sin(2π). Since sin(2π) = 0, then f(π) = π * 0 = 0. Easy enough! For the derivative, f'(x) = sin(2x) + 2x cos(2x), we substitute x with π: f'(π) = sin(2π) + 2π cos(2π). We know that sin(2π) = 0, and cos(2π) = 1. Therefore, f'(π) = 0 + 2π * 1 = 2π. Almost there, guys! We've found f(π) and f'(π). Now, all that is left is to put them together.

We have now found both parts of our original equation. We have found the original function evaluated, and we have the derivative of the original function evaluated. Now we are just going to combine them in the last step. It is as easy as pie! Remember, with practice, these kinds of problems become more intuitive. Keep going, you are doing a great job!

Calculating the Final Answer: The Grand Finale

Finally, let's calculate the final answer: f'(π) + f(π) + 2. We found that f'(π) = 2π and f(π) = 0. So, the expression becomes 2π + 0 + 2, which simplifies to 2π + 2 or 2 + 2π. And that's our answer! We've successfully solved the problem! Congratulations! You see, it wasn't that bad, right?

So, the correct answer is C) 2 + 2π. We've gone from the initial function, found its derivative, evaluated both at a specific point, and then combined them to get the final result. It's a journey, but a rewarding one! Remember to break down complex problems into smaller, manageable steps. That way, you won't get overwhelmed and it makes the entire process way more easier. Keep practicing, and you'll become a calculus pro in no time.

Recap and Key Takeaways

Alright, let's quickly recap what we did and what you should take away from this problem. First, we identified the function and understood that it required the product rule. Then, we found the derivative using the product rule, which is a key skill. After that, we evaluated both the original function and its derivative at the given value of π. Finally, we plugged the results into the given expression and calculated the answer. The key takeaways here are:

  • Product Rule: Always be familiar with the product rule. Recognizing when to apply it is the first step.
  • Chain Rule: If the derivative is of a composite function, make sure you use the chain rule to derive the inside function.
  • Evaluation: Remember how to evaluate functions and their derivatives at specific points.
  • Break it Down: Break the big problem into smaller steps. It is always easier to tackle it that way. You can do it!

This kind of problem is common in calculus courses and beyond. Understanding these concepts will help you build a solid foundation. Keep practicing, and you'll get better and better. Good job, and keep up the great work! You've successfully completed the problem. Keep practicing, and you'll be a calculus wizard in no time. You got this!