Domain Of Composite Functions F(x) = X^2 - 1 And G(x) = 2x - 3

Hey there, math enthusiasts! Today, we're diving into the fascinating world of composite functions and their domains. We'll be tackling a classic problem that involves finding the domain of a composite function, so buckle up and let's get started!

Understanding the Problem: Deconstructing (f ∘ g)(x)

The problem we're tackling today presents us with two functions: f(x) = x² - 1 and g(x) = 2x - 3. The core question is: what's the domain of the composite function (f ∘ g)(x)? This notation, (f ∘ g)(x), might look a bit intimidating at first, but it's simply a way of expressing function composition. Basically, it means we're plugging the entire function g(x) into the function f(x). Think of it like a mathematical assembly line – g(x) does its thing first, and then its output becomes the input for f(x).

To really grasp this, let's break down what (f ∘ g)(x) actually means. It's equivalent to f(g(x)). So, wherever we see 'x' in the function f(x), we're going to replace it with the entire expression for g(x). This is a crucial step in understanding composite functions. We're not just multiplying the functions together; we're feeding one function into another. This process can significantly alter the domain of the resulting composite function, which is precisely what we're trying to figure out today.

Before we jump into the calculations, it's important to remember what a domain actually is. In the simplest terms, the domain of a function is the set of all possible input values (usually 'x' values) for which the function produces a valid output. A valid output is typically a real number. We need to be mindful of situations that might lead to undefined results, such as division by zero or taking the square root of a negative number. These are the usual suspects when we're hunting for domain restrictions.

In the context of composite functions, the domain gets a bit more nuanced. We need to consider not only the domain of the outer function f(x), but also the domain of the inner function g(x), and how the range of g(x) interacts with the domain of f(x). This is where things get interesting, and where we'll need to be extra careful in our analysis. We're not just looking at the individual functions in isolation; we're examining how they work together as a team. This collaborative aspect of composite functions is what makes them so powerful and versatile in mathematics.

So, to summarize, we're on a mission to find all the 'x' values that we can legally plug into (f ∘ g)(x) without causing any mathematical mayhem. We'll need to keep a close eye on both f(x) and g(x), and how they interact with each other. Are you ready to dive in and uncover the secrets of this composite function's domain? Let's do it!

Finding the Composite Function: (f ∘ g)(x) in Action

Okay, let's get our hands dirty and actually find the expression for (f ∘ g)(x). Remember, this means we're plugging g(x) into f(x). We know that f(x) = x² - 1 and g(x) = 2x - 3. So, wherever we see an 'x' in f(x), we're going to replace it with the entire expression 2x - 3. This is the heart of function composition, guys!

So, we start with f(x) = x² - 1. Now, replace 'x' with g(x), which is 2x - 3. This gives us f(g(x)) = (2x - 3)² - 1. See how we've substituted the entire function g(x) into f(x)? This is the key step. Now, we need to simplify this expression to get a clearer picture of what our composite function looks like.

Let's expand the squared term. Remember the binomial expansion: (a - b)² = a² - 2ab + b². Applying this to (2x - 3)², we get (2x)² - 2(2x)(3) + (3)², which simplifies to 4x² - 12x + 9. Don't rush this step – it's easy to make a small arithmetic error that can throw off your entire solution. Double-check your work to make sure you've expanded the square correctly. We want to be precise in our calculations.

Now, let's put it all together. We have f(g(x)) = (2x - 3)² - 1 = 4x² - 12x + 9 - 1. Combining the constant terms, we get f(g(x)) = 4x² - 12x + 8. Boom! We've found our composite function. This quadratic expression, 4x² - 12x + 8, is the result of plugging g(x) into f(x). It's a brand new function, born from the combination of two others.

Now that we have the explicit form of (f ∘ g)(x), we can start thinking about its domain. Remember, the domain is all the possible input values (x-values) that give us a valid output. We need to examine our composite function, 4x² - 12x + 8, and see if there are any restrictions on what 'x' can be. Are there any values that would make the function undefined? That's the question we need to answer. We're on the home stretch now, so let's keep the momentum going!

Determining the Domain: Unveiling the Allowed Inputs

Alright, we've arrived at the crucial part: figuring out the domain of our composite function, (f ∘ g)(x) = 4x² - 12x + 8. Remember, the domain is the set of all possible 'x' values that will produce a real number output. To find the domain, we need to look for any restrictions on 'x'. These restrictions typically arise from two main sources: division by zero and taking the square root (or any even root) of a negative number.

Let's examine our function, 4x² - 12x + 8. Do we see any fractions where the denominator could potentially be zero? Nope. How about any square roots or other even roots? Again, no. This is excellent news! It means that our composite function doesn't have any of the usual suspects that cause domain restrictions. But why is this the case? Let's think about the original functions that made up our composite.

Our original functions were f(x) = x² - 1 and g(x) = 2x - 3. Notice that f(x) is a polynomial (a quadratic, specifically), and g(x) is also a polynomial (a linear function). Polynomials are incredibly well-behaved when it comes to domains. They are defined for all real numbers. You can plug in any 'x' value you can imagine into a polynomial, and you'll always get a real number output. There are no sneaky fractions or square roots to worry about.

Since both f(x) and g(x) have domains of all real numbers, and our composite function (f ∘ g)(x) turned out to be a polynomial as well, we can confidently conclude that the domain of (f ∘ g)(x) is also all real numbers. This means there are no restrictions on 'x'. We can plug in any value, from negative infinity to positive infinity, and our function will happily spit out a real number result. Isn't that satisfying?

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