Domain Of F(x) = 10g(6-3x)

Hey guys, let's dive into the awesome world of algebra and tackle a super common problem: finding the domain of a function. Today, we're going to break down how to find the domain of f(x) = 10g(6-3x). Don't worry if this looks a bit tricky at first; we'll go through it step-by-step, making sure you guys understand every bit of it. Finding the domain is all about figuring out which input values (the 'x' values) are allowed for a function. It's like setting the rules for your function so it doesn't break! When we talk about the domain, we're essentially looking for any restrictions on 'x'. These restrictions usually pop up when we have things like division by zero or when we're taking the square root of a negative number, or even when we're dealing with logarithms. For our specific function, f(x) = 10g(6-3x), the key thing to notice is the 'g' part. In algebra, 'g' often represents a function itself. Without knowing what 'g' is, we can't find a specific numerical domain. However, we can determine the domain based on the argument of the function 'g'. Most commonly, when you see 'g' without a specific definition in this context, it might be implying a generic function, or perhaps it's a placeholder for a common function like a logarithm, as the original prompt might have hinted at with '10g' which could be interpreted as log base 10. Let's assume for a moment that '10g' means the base-10 logarithm function, often written as 'log₁₀'. If that's the case, the function would look more like f(x) = log₁₀(6-3x). Now, the crucial rule for logarithms is that their argument must always be positive. You can't take the logarithm of zero or a negative number in the realm of real numbers. So, for f(x) = log₁₀(6-3x), we need to ensure that the expression inside the logarithm, which is (6-3x), is strictly greater than zero. This inequality, 6 - 3x > 0, is what we need to solve to find the domain. Let's get solving! We want to isolate 'x' on one side of the inequality. First, subtract 6 from both sides: -3x > -6. Now, here's a super important algebra rule to remember: when you divide or multiply both sides of an inequality by a negative number, you must flip the inequality sign. So, when we divide both sides by -3, the '>' sign becomes '<'. This gives us x < (-6) / (-3), which simplifies to x < 2. Therefore, if '10g' implies the base-10 logarithm, the domain of the function f(x) = log₁₀(6-3x) is all real numbers 'x' such that x is less than 2. We can write this in interval notation as (-∞, 2). This means any number from negative infinity up to, but not including, 2 is a valid input for this function. Pretty neat, right? Understanding these fundamental rules, like the conditions for logarithms, is key to mastering function domains. We'll explore other scenarios too, so stick around!

Understanding Function Domains: Why They Matter

Alright guys, let's really get into why finding the domain of a function is such a big deal in algebra. Think of a function like a machine. You put something in (the input, 'x'), and it spits something out (the output, 'f(x)'). The domain is basically the set of all the legitimate things you can put into that machine without breaking it or getting a nonsensical answer. If you try to feed it something it can't handle, like dividing by zero or taking the square root of a negative number, the machine malfunctions – that's what happens when an input is outside the domain. For our specific function, f(x) = 10g(6-3x), the core of finding its domain lies in understanding what the '10g' part means and what restrictions that places on the argument (6-3x). As we discussed, if '10g' is interpreted as the base-10 logarithm (log₁₀), then the argument must be positive. This is a fundamental property of logarithms: you can only take the log of a positive number. Why? Well, think about what logarithms actually do. The expression log_b(a) = c asks, 'To what power do I need to raise the base 'b' to get the number 'a'?'. For example, log₁₀(100) = 2 because 10² = 100. If you try to find log₁₀(0), you're asking '10 to what power equals 0?'. There's no real number power that works. Similarly, for log₁₀(-10), you're asking '10 to what power equals -10?'. Again, no real number power of a positive base (like 10) can ever result in a negative number. That's why the argument has to be positive. So, we set up the inequality 6 - 3x > 0. Let's solve this step-by-step, making sure we handle the inequality rules correctly. First, we want to isolate the term with 'x'. Subtract 6 from both sides: -3x > -6. Now, this is the critical step: we need to divide by -3 to get 'x' by itself. Remember the golden rule of inequalities: when you multiply or divide by a negative number, you must flip the direction of the inequality sign. So, the '>' sign flips to '<'. Dividing -6 by -3 gives us 2. Thus, we get x < 2. This means that for the function f(x) = log₁₀(6-3x) to produce a real number output, the input 'x' must be any number less than 2. Graphically, this represents a ray starting from negative infinity and extending up to, but not including, the number 2 on the number line. In interval notation, this is written as (-∞, 2). This notation tells us the range of allowed 'x' values. It's super important to be comfortable with both inequality notation (x < 2) and interval notation (-∞, 2). If '10g' represented something else, like perhaps a typo and it should have been a different function, the process might change. For instance, if it was a simple linear function like g(x) = 6 - 3x, then f(x) = 10 * (6 - 3x), which is just a linear function, and linear functions have a domain of all real numbers (-∞, ∞) because you can plug in any 'x' value and get a valid output. But given the structure and the common interpretations in algebra problems, the logarithm interpretation is the most likely intended scenario. Always pay attention to the specific functions involved and their inherent restrictions.

Breaking Down the Inequality: 6 - 3x > 0

Let's zoom in on the core mathematical operation that dictates the domain for our function f(x) = 10g(6-3x), assuming '10g' implies a base-10 logarithm. The critical part is the argument of the logarithm, which is 6 - 3x. As we've stressed, for any logarithm function, say log_b(argument), the argument must always be strictly greater than zero. This is a non-negotiable rule in the world of real numbers. So, we translate this rule directly into an inequality: 6 - 3x > 0. Our mission now is to solve this inequality for 'x', which will tell us exactly which values of 'x' are permissible inputs for our function. We're essentially setting the bounds for our function's domain. Think of it as finding the 'safe zone' for 'x'. Let's walk through the algebraic steps. Our inequality is 6 - 3x > 0. The first step is usually to get the term containing 'x' by itself on one side. We can do this by subtracting 6 from both sides of the inequality. Remember, whatever you do to one side, you must do to the other to keep the inequality balanced. So, 6 - 3x - 6 > 0 - 6, which simplifies to -3x > -6. Now, we're one step away from isolating 'x'. We need to get rid of the '-3' coefficient. To do this, we divide both sides of the inequality by -3. And here comes the most important rule for inequalities: Whenever you multiply or divide an inequality by a negative number, you absolutely MUST reverse the direction of the inequality sign. This is a common tripping point for many students, so always keep it in mind! Our '>' sign needs to flip to a '<' sign. So, dividing both sides by -3, we get: (-3x) / (-3) < (-6) / (-3). On the left side, the -3's cancel out, leaving us with just 'x'. On the right side, -6 divided by -3 equals positive 2. So, the inequality becomes x < 2. This result, x < 2, is the solution to our inequality. It tells us that any value of 'x' that is less than 2 will make the argument (6 - 3x) positive, thus allowing the logarithm to be calculated. Conversely, if x is 2 or greater, the argument will be zero or negative, and the logarithm would be undefined in the real number system. Therefore, the domain of our function f(x) = log₁₀(6-3x) consists of all real numbers less than 2. We can express this domain using interval notation as (-∞, 2). This interval includes all numbers from negative infinity up to, but not including, 2. It's a crucial step in understanding the behavior and limitations of mathematical functions.

Expressing the Domain: Inequality and Interval Notation

Now that we've done the hard work of solving the inequality for our function f(x) = 10g(6-3x) (assuming '10g' is the base-10 logarithm), the final step is to clearly express the domain. Guys, it's super important to be able to communicate your answer effectively, and in mathematics, we have specific ways to denote domains. We found that the condition for the function to be defined is x < 2. This is known as inequality notation. It directly tells us the relationship between 'x' and the boundary value 2. It means 'x' can be any real number that is strictly less than 2. However, often in algebra and calculus, we use interval notation because it's concise and very descriptive, especially when dealing with more complex domains. To convert x < 2 into interval notation, we think about the number line. The numbers that satisfy x < 2 start from the far left (negative infinity) and go all the way up to the number 2. Since 'x' must be strictly less than 2, the number 2 itself is not included in the domain. In interval notation, we use parentheses () to indicate that an endpoint is not included, and square brackets [] to indicate that an endpoint is included. Infinity symbols (-∞ and ∞) are always used with parentheses because you can never reach or include infinity. So, for x < 2, the interval starts at negative infinity (-∞), which always gets a parenthesis, and ends at 2. Since 2 is not included, we use another parenthesis ). Putting it all together, the interval notation for our domain is (-∞, 2). This notation is read as 'the interval from negative infinity to 2, exclusive'. It encompasses all possible real number inputs for 'x' that will result in a valid, defined output for the function f(x) = log₁₀(6-3x). Understanding both inequality notation (like x < 2) and interval notation (like (-∞, 2)) is essential for your math journey. They are just two different languages to describe the same set of numbers, and knowing both will make you a more versatile mathematician. Keep practicing, and you'll master these in no time!

Conclusion: Mastering Function Domains

So there you have it, guys! We've successfully navigated the process of finding the domain for the function f(x) = 10g(6-3x), making the crucial assumption that '10g' refers to the base-10 logarithm. The key takeaway is that the domain of a function is determined by any constraints imposed by its components. For logarithms, this constraint is fundamental: the argument must always be positive. By setting the argument (6 - 3x) to be greater than zero (6 - 3x > 0) and solving the resulting inequality, we uncovered that x < 2. This means that only values of 'x' less than 2 are valid inputs for this function to produce a real number output. We've expressed this domain in both inequality notation (x < 2) and interval notation ((-∞, 2)). Remember this process: identify potential restrictions (like denominators or arguments of logs/roots), set up the necessary inequalities or conditions, and solve for 'x'. Mastering domain calculations is a cornerstone of algebra and will serve you well as you tackle more complex mathematical concepts. Keep practicing these skills, and you'll become a domain-finding pro! If '10g' was intended to represent something else, the domain calculation would indeed change, highlighting the importance of precise notation and understanding of different mathematical functions. Always double-check the definitions provided or implied in your problems. Happy problem-solving!