Decomposing Composite Functions Finding A And B In H(x) = (x-1)³ + 10

Jul 24, 2025 by ADMIN 70 views

Hey guys! Let's dive into the fascinating world of composite functions. Today, we're going to tackle a problem where we need to break down a composite function and figure out the values of some unknown constants. Specifically, we're looking at the composite function h(x) = (x - 1)³ + 10 and trying to find two functions, f(x) and g(x), that when composed, give us h(x). The functions f(x) and g(x) have specific forms: f(x) = x + a and g(x) = x³ + b. Our mission, should we choose to accept it (and we do!), is to find the values of a and b that make this composition work.

Understanding Composite Functions

Before we jump into the nitty-gritty, let's quickly recap what composite functions are all about. A composite function is essentially a function within a function. Think of it like a set of Russian nesting dolls, where one function's output becomes the input for another. Mathematically, we represent the composition of g with f as (g ∘ f)(x), which means we first apply the function f to x, and then we apply the function g to the result. So, (g ∘ f)(x) = g(f(x)). This notation is super important, so make sure you're comfortable with it. Composite functions are the bread and butter of advanced mathematics, allowing us to model complex relationships by chaining together simpler operations. The key idea is that the output of the inner function becomes the input of the outer function, creating a flow of operations that ultimately defines the composite function's behavior. In our case, we're given the final composite function h(x) and the general forms of the inner and outer functions f(x) and g(x). Our goal is to reverse-engineer the composition process, figuring out the specific values of a and b that make the composition work. This kind of problem is a classic example of function decomposition, where we break down a complex function into its simpler building blocks. Mastering this skill is essential for understanding more advanced mathematical concepts and applying them to real-world problems. Now that we're all on the same page about composite functions, let's get back to our specific problem and start figuring out how to find those values of a and b. Remember, the goal is to make the composition g(f(x)) equivalent to the given function h(x). This will involve some algebraic manipulation and careful comparison of coefficients, but don't worry, we'll break it down step by step.

Setting Up the Composition

Okay, let's get our hands dirty with some algebra. We know that h(x) = (g ∘ f)(x), which means h(x) = g(f(x)). We're given that f(x) = x + a and g(x) = x³ + b. So, to find g(f(x)), we need to substitute f(x) into g(x) wherever we see an x. This gives us g(f(x)) = g(x + a) = (x + a)³ + b. Now, we have an expression for the composite function in terms of a and b. The next step is to expand this expression and compare it to the given expression for h(x). Setting up the composition correctly is crucial for solving this problem. If we make a mistake in this initial step, it will throw off the rest of our calculations. That's why it's important to be careful and double-check our work. We're essentially building a bridge between the given functions f(x) and g(x) and the target function h(x). The composition operation is the tool we use to construct this bridge. By substituting f(x) into g(x), we're creating a new function that incorporates the effects of both f and g. This is the essence of function composition: combining functions to create more complex and interesting behaviors. In this specific case, we're using the composition to link two simple functions, a linear function f(x) and a cubic function g(x), to a slightly more complex cubic function h(x). The parameters a and b act as the adjustable knobs that allow us to fine-tune the composition and make it match h(x) exactly. Now that we've successfully set up the composition, the next step is to unleash our algebraic skills and expand the expression (x + a)³ + b. This will involve using the binomial theorem or simply multiplying out the factors, but either way, we'll end up with a polynomial expression that we can then compare to the given expression for h(x).

Expanding and Comparing

Alright, time to expand (x + a)³ + b. Remember the binomial theorem, or you can just multiply it out step by step: (x + a)³ = (x + a)(x + a)(x + a) = (x² + 2ax + a²)(x + a) = x³ + 3ax² + 3a²x + a³. So, g(f(x)) = x³ + 3ax² + 3a²x + a³ + b. Now, we know that h(x) = (x - 1)³ + 10. Let's expand that too: h(x) = (x - 1)(x - 1)(x - 1) + 10 = (x² - 2x + 1)(x - 1) + 10 = x³ - 3x² + 3x - 1 + 10 = x³ - 3x² + 3x + 9. Now we have two expressions: g(f(x)) = x³ + 3ax² + 3a²x + a³ + b and h(x) = x³ - 3x² + 3x + 9. For these two functions to be equal, their corresponding coefficients must be equal. This is a crucial step in solving the problem. We're essentially saying that if two polynomials are equal, then the coefficients of the terms with the same degree must be equal. This is a fundamental property of polynomials and allows us to set up a system of equations to solve for our unknowns. Expanding and comparing the expressions is where the rubber meets the road. We're taking the abstract idea of function composition and turning it into a concrete algebraic problem. By expanding the expressions, we're revealing the underlying structure of the polynomials and making it possible to compare them term by term. This process of expanding and comparing is not just a mathematical trick; it's a way of understanding how the parameters a and b affect the shape and position of the composite function. Each coefficient in the expanded polynomial corresponds to a specific feature of the function's graph, such as its curvature, slope, and vertical position. By matching the coefficients, we're ensuring that the composite function g(f(x)) has the same features as the target function h(x). Now that we have the expanded expressions and understand the principle of comparing coefficients, the next step is to actually set up the equations and solve for a and b. This will involve some careful manipulation of the equations, but we're well on our way to finding the solution. Remember, the key is to match the coefficients of the corresponding terms in the two polynomials. This will give us a system of equations that we can then solve using standard algebraic techniques.

Solving for a and b

Let's equate the coefficients. We have:

Coefficient of x²: 3a = -3
Coefficient of x: 3a² = 3
Constant term: a³ + b = 9

From the first equation, 3a = -3, we get a = -1. Let's check if this value satisfies the second equation, 3a² = 3. Substituting a = -1, we get 3(-1)² = 3, which is true! Awesome! Now, let's use the third equation, a³ + b = 9, to find b. Substituting a = -1, we get (-1)³ + b = 9, which simplifies to -1 + b = 9. Adding 1 to both sides, we get b = 10. So, we found our values: a = -1 and b = 10. Solving for a and b is the final step in our algebraic journey. We've set up the problem, expanded the expressions, compared the coefficients, and now we're ready to reap the rewards of our hard work. The equations we've obtained represent the mathematical constraints that ensure the composite function g(f(x)) matches the target function h(x). Each equation tells us something specific about the relationship between a, b, and the coefficients of the polynomials. By solving these equations, we're essentially unlocking the values of a and b that make the composite function puzzle fit together perfectly. The process of solving a system of equations is a fundamental skill in mathematics and has wide-ranging applications in science, engineering, and economics. It's a powerful tool for finding the unknown parameters that govern a system or process. In this case, we're using it to find the values of a and b that define the functions f(x) and g(x) and make their composition equal to h(x). Now that we've found the values of a and b, it's always a good idea to check our work and make sure that our solution is correct. We can do this by plugging the values back into the original equations or by directly computing the composite function g(f(x)) with our found values of a and b and verifying that it is indeed equal to h(x).

Verification

Let's verify our solution. We found a = -1 and b = 10. So, f(x) = x - 1 and g(x) = x³ + 10. Now, let's compute g(f(x)) = g(x - 1) = (x - 1)³ + 10. Hey, that's exactly h(x)! We did it! We successfully decomposed the composite function and found the values of a and b. Verification is the crucial final step in any mathematical problem. It's our way of ensuring that our solution is not just a lucky guess but a true and accurate answer. By plugging our values of a and b back into the original problem, we're putting our solution to the test. If the verification fails, it means we've made a mistake somewhere along the way and need to go back and re-examine our work. But if the verification succeeds, as it does in this case, it gives us confidence that we've solved the problem correctly. The verification process also reinforces our understanding of the problem and the concepts involved. By explicitly computing the composite function with our found values of a and b, we're solidifying our grasp of function composition and how the parameters a and b influence the overall behavior of the composite function. In this specific problem, the verification step is particularly satisfying because it brings us full circle. We started with the composite function h(x), broke it down into its constituent functions f(x) and g(x), and then reconstructed h(x) by composing f(x) and g(x) with our found values of a and b. This circular process is a beautiful illustration of the power and elegance of mathematical reasoning.

Conclusion

So, there you have it! We found that a = -1 and b = 10 are the values that make the composition h(x) = (g ∘ f)(x) true, given f(x) = x + a and g(x) = x³ + b. This problem highlights the power of understanding composite functions and how we can manipulate them to solve for unknowns. It's like being a mathematical detective, piecing together clues to solve the puzzle. In conclusion, we've successfully navigated the world of composite functions and found the values of a and b that satisfy the given conditions. This problem is a great example of how algebraic manipulation and careful comparison of coefficients can be used to solve complex mathematical problems. The key takeaways from this exercise are the importance of understanding function composition, the power of expanding and comparing polynomial expressions, and the necessity of verifying our solutions. By mastering these skills, we can tackle a wide range of mathematical problems and gain a deeper appreciation for the beauty and elegance of mathematics. The process we followed in this problem, from setting up the composition to expanding and comparing expressions to solving for the unknowns and verifying our solution, is a general approach that can be applied to many other similar problems. It's a powerful toolkit for any aspiring mathematician or problem solver. And remember, the journey of solving a mathematical problem is just as important as the final answer. It's through the process of grappling with the problem, making mistakes, learning from them, and ultimately finding the solution that we truly develop our mathematical skills and intuition. So keep practicing, keep exploring, and keep having fun with mathematics! You guys rock!