Understanding Function G(x) = 2 + 3/x: A Full Guide

Hey math lovers! Today, we're diving deep into a super interesting function: g(x) = 2 + 3/x. This isn't just any old equation; it's a rational function, and understanding these guys is key to unlocking more complex math concepts. We'll be breaking down how to find specific values, solve for unknowns, and even explore when a function maps to itself. So grab your notebooks, maybe a cup of your favorite beverage, and let's get this math party started!

Part (i): Finding the Value of k

Alright, first things first, let's talk about this 'k' in our function definition, g(x) = 2 + 3/x, with x ≠ k. You see that little restriction, x ≠ k? That's a big clue! In rational functions, there's often a value that the variable (in this case, 'x') cannot be. Why? Because it would make the function undefined, usually by causing division by zero. Think about it: you can't divide by zero in math; it's like trying to split a pizza among zero people – it just doesn't make sense! So, for our function g(x) = 2 + 3/x, the part that could potentially cause trouble is the 3/x term. If 'x' were equal to zero, we'd be dividing 3 by zero. Boom! Undefined. Therefore, to keep our function happy and well-behaved, we must state that x cannot be 0. So, the value of k is 0. It's that simple! This restriction is super important because it tells us the domain of the function – all the possible x-values we can plug in. Knowing the domain helps us avoid errors and understand the function's behavior better. When you're working with rational functions, always be on the lookout for what makes the denominator zero. That value is your 'k', your forbidden zone for 'x'. It's like a secret code to understanding where the function lives and breathes!

Part (ii): Calculating g(-2)

Now that we've got the hang of our function g(x) = 2 + 3/x and know that x ≠ 0, let's find out what happens when we plug in a specific number. The question asks us to find g(-2). This is like asking, "What's the output of the machine if we put in -2?". To do this, we simply substitute '-2' for every 'x' in our function's formula. So, we have: g(-2) = 2 + 3/(-2). First, let's handle the fraction: 3 divided by -2 is -1.5. Now, we add that to the 2. So, g(-2) = 2 + (-1.5). And voilà! g(-2) = 0.5. See? It's just a substitution game! This step is fundamental in understanding functions. It shows us how specific inputs lead to specific outputs. For every valid 'x' you input, you get a unique 'g(x)' value. We're essentially tracing a point on the graph of this function. The point (-2, 0.5) is part of the graph of g(x). Pretty neat, right? This process of substituting values is also how we check our work and how we'll approach the next parts of this problem. It reinforces the idea that functions are like rulebooks; you follow the rules, and you get a result.

Part (iii): Solving for x when g(x) = -6

Okay, guys, this is where things get a little more exciting! We're not just plugging in numbers anymore; we're working backward. The question is: find the value of x when g(x) = -6. This means we know the output of our function is -6, and we need to discover the input that produced it. So, we set our function equal to -6: 2 + 3/x = -6. Our goal now is to isolate 'x'. First, let's subtract 2 from both sides of the equation to get the fraction by itself: 3/x = -6 - 2, which simplifies to 3/x = -8. Now, to get 'x' out of the denominator, we can multiply both sides by 'x': 3 = -8x. Finally, to find 'x', we divide both sides by -8: x = 3 / -8, or x = -3/8. So, when our function g(x) is equal to -6, the input value x is -3/8. This process of solving for 'x' when given a specific output is crucial. It helps us understand the inverse relationship within functions and is a core skill in algebra. We're essentially reversing the function's operation. If g(x) = -6, then x = -3/8. We can even check this: g(-3/8) = 2 + 3/(-3/8) = 2 + 3 * (-8/3) = 2 + (-8) = -6. It works out perfectly! This confirms our calculation and demonstrates the power of algebraic manipulation.

Part (iv): Finding x values that Map to Themselves

This last part is super cool and often trips people up, but it's totally manageable once you get the concept. We need to find the values of x which map to themselves. What does that even mean, right? It means we're looking for an 'x' value such that when you plug it into the function g(x), the output is the exact same as the input. In other words, g(x) = x. So, we set our function equal to 'x': 2 + 3/x = x. Now, we need to solve this equation for 'x'. Just like before, we want to get rid of that pesky fraction. Multiply the entire equation by 'x' (remembering that x ≠ 0): x(2 + 3/x) = x(x). This gives us 2x + 3 = x². Now, this looks like a quadratic equation, so let's rearrange it into the standard form ax² + bx + c = 0. We can do this by subtracting 2x and 3 from both sides: 0 = x² - 2x - 3. Okay, deep breaths! We can solve this quadratic equation by factoring. We need two numbers that multiply to -3 and add up to -2. Those numbers are -3 and +1. So, we can factor the equation as (x - 3)(x + 1) = 0. For this product to be zero, one of the factors must be zero. So, either x - 3 = 0 (which means x = 3) or x + 1 = 0 (which means x = -1). These are our potential solutions! Let's check them. If x = 3, then g(3) = 2 + 3/3 = 2 + 1 = 3. So, g(3) = 3. It maps to itself! Awesome. If x = -1, then g(-1) = 2 + 3/(-1) = 2 + (-3) = -1. So, g(-1) = -1. It maps to itself too! Fantastic! Therefore, the values of x that map to themselves for the function g(x) = 2 + 3/x are 3 and -1. This concept of finding fixed points (where f(x) = x) is super important in many areas of math, from calculus to computer science. It's like finding the 'stable points' of a system. So, we've successfully navigated through finding restrictions, evaluating the function, solving for inputs, and even finding fixed points. Keep practicing these skills, and you'll be a function-solving pro in no time!