Finding Inverse Functions And Compositions For G And H

Hey guys! Today, we're diving into the fascinating world of inverse functions and function composition! We've got two functions to play with: a set of ordered pairs defining g and a linear function h(x). Our mission? To find g⁻¹(5), h⁻¹(x), and (h ∘ h⁻¹)(-5). Let's break it down step by step!

Determining g⁻¹(5)

Let's start with finding the value of g⁻¹(5). Remember, the inverse function, denoted as g⁻¹(x), essentially undoes what the original function g(x) does. If g(a) = b, then g⁻¹(b) = a. This is the core concept we'll use to unravel this problem. The function g is given as a set of ordered pairs: g = {(-5,7), (-4,8), (3,5), (5,2), (7,6)}. Each pair represents an input and its corresponding output. For example, the pair (3, 5) tells us that g(3) = 5. To find g⁻¹(5), we need to look for the ordered pair in g where the output is 5. Scanning through the pairs, we find (3, 5). This means that when the input to g is 3, the output is 5. Therefore, the inverse function g⁻¹ will take the output 5 and return the original input 3. So, we can confidently say that g⁻¹(5) = 3. This concept of reversing the input-output relationship is fundamental to understanding inverse functions. It's like having a machine that turns apples into juice; the inverse function is like having a machine that (hypothetically!) turns the juice back into an apple. Now, let’s solidify this concept with an analogy. Imagine you have a lock and a key. The function g is like locking something with the key, and the function g⁻¹ is like unlocking it. To find g⁻¹(5), we're essentially asking: “What was the input that g used to produce the output 5?” By looking at the set of ordered pairs, we found that the input 3 produced the output 5, so the 'key' to unlocking 5 is 3. This highlights the one-to-one nature of the function g. For each input, there is a unique output, and vice versa. This is a crucial property for a function to have an inverse. If there were two different inputs that produced the same output, the inverse function would be ambiguous – it wouldn't know which input to return. The given set of ordered pairs for g satisfies this condition, ensuring that g⁻¹ is well-defined. So, to recap, finding g⁻¹(5) involves looking at the ordered pairs of g and identifying the pair where the second element (the output) is 5. The first element of that pair (the input) is then the value of g⁻¹(5). In this case, it's a straightforward process of pattern matching and understanding the core concept of inverse functions. Remember, inverse functions are all about reversing the operation, and with a clear understanding of this principle, problems like this become much easier to tackle. Keep practicing and visualizing these concepts, and you'll become a pro at handling inverse functions!

Finding h⁻¹(x)

Next up, let's tackle finding h⁻¹(x). Unlike g, which is defined by a set of points, h is defined by an equation: h(x) = 2x - 3. To find the inverse function h⁻¹(x), we need to reverse the operations that h performs on x. In other words, we need to isolate x in terms of h(x). Here's how we can do it: First, let's replace h(x) with y for simplicity: y = 2x - 3. Now, our goal is to get x by itself on one side of the equation. To do this, we'll perform the inverse operations in reverse order. The first operation that h performs is multiplication by 2, and then it subtracts 3. So, to undo these operations, we'll first add 3 to both sides of the equation: y + 3 = 2x. Next, we'll divide both sides by 2: (y + 3) / 2 = x. Now we have x isolated! This expression tells us how to get the input x back if we know the output y. To write the inverse function in standard notation, we swap x and y: y = (x + 3) / 2. Finally, we replace y with h⁻¹(x) to denote the inverse function: h⁻¹(x) = (x + 3) / 2. So, we've successfully found the inverse function of h(x). Let’s think about what we just did in a more intuitive way. The original function h(x) takes an input x, multiplies it by 2, and then subtracts 3. The inverse function h⁻¹(x) needs to undo these operations in the reverse order. So, it first adds 3 to the input and then divides by 2. This 'undoing' process is the essence of finding inverse functions. This can be thought of as reversing a recipe. If a recipe tells you to first mix ingredients A and B, then bake at a certain temperature, the inverse process would be to 'unbake' (hypothetically!) and then separate ingredients A and B. Of course, in the real world, you can't unbake a cake, but this analogy helps to visualize how inverse functions work mathematically. Now, let's test our understanding with an example. Suppose we want to find the value of h⁻¹(5). We can plug 5 into our expression for h⁻¹(x): h⁻¹(5) = (5 + 3) / 2 = 8 / 2 = 4. This tells us that if h(4) should equal 5. Let's verify this: h(4) = 2(4) - 3 = 8 - 3 = 5. It checks out! This confirms that we have correctly found the inverse function. To summarize, finding the inverse of a function defined by an equation involves swapping x and y and then solving for y. This process effectively reverses the operations performed by the original function. With practice, you'll become fluent in this process and be able to find inverse functions with ease.

Evaluating (h ∘ h⁻¹)(-5)

Lastly, let's determine the value of (h ∘ h⁻¹)(-5). This notation represents the composition of functions, specifically h composed with its inverse h⁻¹. Function composition means applying one function to the result of another. In this case, we're first applying h⁻¹ to -5, and then applying h to the result. A key property of inverse functions is that when you compose a function with its inverse (in either order), you get back the original input. In other words, h(h⁻¹(x)) = x and h⁻¹(h(x)) = x. This is because the inverse function undoes the original function, effectively canceling out their effects. Therefore, we can directly apply this property to our problem. Since (h ∘ h⁻¹)(-5) means h(h⁻¹(-5)), and h and h⁻¹ are inverses, we know that the result will simply be the original input, which is -5. So, (h ∘ h⁻¹)(-5) = -5. That was a quick one, wasn't it? Let's think about why this works intuitively. Imagine you have a machine that performs a series of operations on an input, and then you have a second machine that performs the exact opposite operations in reverse order. If you feed an input into the first machine and then feed the output into the second machine, you'll end up with the original input. The functions h and h⁻¹ are like these two machines. h does something to the input, and h⁻¹ undoes it. Composing them results in no net change to the input. Now, let's consider a slightly more complex example to solidify our understanding. Suppose we had a different function, say f(x) = x², and we wanted to find (f ∘ f⁻¹)(2). We would first need to find f⁻¹(x). However, there's a catch! The function f(x) = x² is not one-to-one over its entire domain (because both positive and negative inputs can produce the same output). This means it doesn't have a true inverse over its entire domain. We would need to restrict the domain of f(x) to make it one-to-one (for example, by only considering non-negative inputs) before we could find a well-defined inverse. In our original problem, h(x) is a linear function, which is always one-to-one, so we don't have this issue. But it's important to be aware of this when dealing with other types of functions. To summarize, the composition of a function with its inverse always results in the original input, provided that the function has a well-defined inverse over the relevant domain. This is a powerful property that simplifies many calculations involving inverse functions. Remember this key concept, and you'll be able to tackle similar problems with confidence!

Conclusion

Alright guys, we've successfully navigated the world of inverse functions and composition! We found that g⁻¹(5) = 3, h⁻¹(x) = (x + 3) / 2, and (h ∘ h⁻¹)(-5) = -5. Remember the core concepts: inverse functions reverse the input-output relationship, and composing a function with its inverse results in the original input. Keep practicing, and you'll master these concepts in no time! Now you're equipped to tackle more challenging problems involving functions and their inverses. Keep up the great work!"