Function Operations: Find (r+s)(x), (r-s)(x), And (r⋅s)(-1)
Hey guys! Today, we're diving into the world of function operations. We'll be working with two functions, specifically r(x) = x² and s(x) = 3x³, and we're going to figure out how to combine them using addition, subtraction, and multiplication. We'll calculate (r + s)(x), (r - s)(x), and finally, evaluate (r ⋅ s)(-1). So, let's get started and make math fun!
Understanding Function Operations
Before we jump into the calculations, let's quickly recap what function operations actually mean. When we talk about (r + s)(x), (r - s)(x), and (r ⋅ s)(x), we're essentially describing how to combine the outputs of the individual functions r(x) and s(x) for a given input x. Function operations are like the basic arithmetic operations but applied to functions. Just as we can add, subtract, multiply, and divide numbers, we can perform these operations on functions as well. This concept is super useful in various areas of mathematics, especially calculus and analysis, as well as fields like physics and engineering where functions are used to model real-world phenomena. By understanding how functions interact through these operations, we can build more complex models and gain deeper insights into the systems we are studying. It is important to grasp these operations well, as they form the cornerstone for advanced mathematical concepts. Let's break down each operation:
- Addition (r + s)(x): This means we add the outputs of the two functions. So, (r + s)(x) = r(x) + s(x).
- Subtraction (r - s)(x): Here, we subtract the output of s(x) from the output of r(x). Hence, (r - s)(x) = r(x) - s(x).
- Multiplication (r ⋅ s)(x): This operation involves multiplying the outputs of r(x) and s(x), expressed as (r ⋅ s)(x) = r(x) * s(x).
Now that we have a solid understanding of these operations, we can move on to applying them to our specific functions.
Calculating (r + s)(x)
Okay, let's start with finding (r + s)(x). Remember, this means we need to add the two functions r(x) and s(x) together. We know that r(x) = x² and s(x) = 3x³. So, to find (r + s)(x), we simply add these two expressions:
(r + s)(x) = r(x) + s(x) = x² + 3x³
That's it! There's not much more we can simplify here since x² and 3x³ are not like terms. They have different exponents, so we can't combine them any further. The final expression for (r + s)(x) is just the sum of the two functions, neatly written as:
(r + s)(x) = x² + 3x³
It’s important to keep the terms separate unless they have the same variable and exponent. This simple addition is a fundamental operation, and mastering it sets the stage for tackling more complex function manipulations later on. Guys, this is just the beginning, so keep your thinking caps on as we move to the next operation.
Finding (r - s)(x)
Next up, we're going to calculate (r - s)(x). This time, instead of adding the functions, we're going to subtract s(x) from r(x). Again, we have r(x) = x² and s(x) = 3x³. To find (r - s)(x), we subtract s(x) from r(x):
(r - s)(x) = r(x) - s(x) = x² - 3x³
Just like with addition, we can't simplify this expression any further because x² and 3x³ are not like terms. The exponents are different, so they can't be combined. So, our final expression for (r - s)(x) is:
(r - s)(x) = x² - 3x³
Subtraction, in the context of function operations, is a crucial way to compare and contrast how two functions behave relative to each other. The order of subtraction matters here; (r - s)(x) is different from (s - r)(x). In this case, we're seeing how r(x) changes when we take away s(x). Guys, understanding this difference is key to working with more advanced concepts in calculus and beyond.
Evaluating (r ⋅ s)(-1)
Now, let's tackle the multiplication of the functions and then evaluate the result at a specific point. We need to find (r ⋅ s)(-1). This means we first need to find the expression for (r ⋅ s)(x), which is r(x) * s(x), and then substitute x = -1 into that expression.
We know that r(x) = x² and s(x) = 3x³. So, let's multiply these two functions together:
(r ⋅ s)(x) = r(x) * s(x) = x² * (3x³)
When we multiply these terms, we multiply the coefficients and add the exponents:
(r ⋅ s)(x) = 3x^(2+3) = 3x⁵
Now that we have the expression for (r ⋅ s)(x), we can substitute x = -1 to find (r ⋅ s)(-1):
(r ⋅ s)(-1) = 3(-1)⁵
Remember that a negative number raised to an odd power is negative:
(-1)⁵ = -1
So,
(r ⋅ s)(-1) = 3 * (-1) = -3
Therefore, the value of (r ⋅ s)(-1) is:
(r ⋅ s)(-1) = -3
This final evaluation combines function multiplication with numerical substitution, a common technique in various mathematical applications. Guys, multiplying functions and then evaluating them helps us understand how the combined function behaves at different points, providing insights into its overall characteristics and behavior.
Conclusion
Awesome! We've successfully navigated through function operations today. We started with the functions r(x) = x² and s(x) = 3x³ and figured out how to combine them using addition, subtraction, and multiplication. We found that:
- (r + s)(x) = x² + 3x³
- (r - s)(x) = x² - 3x³
- (r ⋅ s)(-1) = -3
Understanding these operations is super important because they pop up everywhere in math, from calculus to more advanced topics. By grasping the basics of adding, subtracting, and multiplying functions, you're building a solid foundation for all sorts of mathematical adventures ahead. Guys, function operations are not just abstract concepts; they are the building blocks for modeling real-world phenomena, solving complex equations, and much more. So keep practicing, keep exploring, and remember that every mathematical challenge is an opportunity to learn and grow. Great job today, and keep up the amazing work! Remember, math is not just about getting the right answers; it’s about understanding the process and building a logical way of thinking. So next time you encounter a similar problem, take a deep breath, break it down step by step, and you'll be solving it like a pro in no time. Keep the questions coming, and let's keep learning together!