Function Evaluation: Unveiling F(x) And Its Domain

Hey guys, let's dive into some math! We're gonna explore a cool function, f(x) = sqrt(x-2) / (x^2 - 8), and figure out how it behaves with different inputs. This is all about function evaluation, which is super important in understanding how functions work. We'll be looking at what happens when you plug in specific values for x and see if we get a real number, or if things get a bit wonky. Remember, understanding functions is like having a superpower – it helps you unlock so many other math concepts! Let's get started and make sure we fully grasp this, ensuring you're confident in evaluating functions.

(a) Evaluate f(21)

Alright, let's start with part (a) and evaluate f(21). This means we need to substitute 21 for x in our function. So, we'll replace every x in f(x) = sqrt(x-2) / (x^2 - 8) with 21. This gives us f(21) = sqrt(21-2) / (21^2 - 8). Now, let's simplify this step by step. First, calculate the value inside the square root: 21 - 2 = 19. Next, calculate the square of 21: 21^2 = 441. And finally, subtract 8: 441 - 8 = 433. Putting it all together, we have f(21) = sqrt(19) / 433. The square root of 19 is a real number, and 433 is a non-zero number. Thus, we end up with a defined value. The result f(21) = sqrt(19) / 433 is a valid result. Therefore, f(21) is defined, and our simplified result is sqrt(19) / 433. Pretty straightforward, right? This part highlights the importance of order of operations! Always remember the proper order to avoid mistakes, as it's an important aspect of function evaluation and finding accurate results.

Now, let's analyze the options: A. f(21) = sqrt(19) / 433 This is the correct calculation and is defined. B. f(21) is undefined. This is incorrect because we arrived at a defined value. So, the correct answer here is A. Evaluating f(21) involves substituting the value and carefully simplifying the expression. There is no issue with the square root because the number inside it is positive, and the denominator is not zero.

(b) Evaluate f(-6)

Now, let’s move on to part (b) and figure out f(-6). We follow the same process, but this time we substitute -6 for x. This gives us f(-6) = sqrt(-6 - 2) / ((-6)^2 - 8). First, let’s simplify the expression inside the square root: -6 - 2 = -8. This is where we run into a potential problem. We end up with the square root of a negative number. The square root of a negative number is not a real number; it is an imaginary number. For our purposes, we'll consider it undefined in the real number system.

Let’s finish evaluating the rest of the function to see how it looks if we ignore the square root issue to emphasize the importance of it. Then, calculate the square of -6: (-6)^2 = 36. Next, subtract 8: 36 - 8 = 28. So, we would have sqrt(-8) / 28 if we could proceed. However, since we can't take the square root of -8 in the real numbers, this operation is undefined. In essence, the domain of a function can give the restrictions on what values are allowed in the function, such as the square root. Thus, the correct answer is B. f(-6) is undefined. Remember, the square root of a negative number is not a real number, meaning it is undefined in the real number domain.

Let's analyze the options: A. f(-6) = This implies there's a real number answer. B. f(-6) is undefined. This is correct because we arrived at a square root of a negative number. This part highlights the domain restrictions imposed by the square root and how it makes the function undefined for specific inputs. Always be aware of the restrictions when you're evaluating the function and working with the domain.

(c) A Deeper Dive into Domain and Function Behavior

Alright, guys, let's take a step back and talk more generally about the domain of a function and how it relates to what we've been doing. The domain is essentially the set of all possible input values (x-values) for which the function produces a valid output (a real number in this case). When evaluating functions, you need to be mindful of values that are excluded from the domain because they lead to undefined results. For our function f(x) = sqrt(x-2) / (x^2 - 8), the domain is determined by two main considerations: the square root and the denominator.

First, consider the square root: sqrt(x-2). The expression inside the square root must be greater than or equal to zero (non-negative) to have a real result. Therefore, we must have x - 2 >= 0, which means x >= 2. This tells us that any x-value less than 2 is not in the domain because the square root would be of a negative number. For example, in our evaluation of f(-6), we saw that x = -6 leads to a negative value inside the square root, making the function undefined at that point. Second, let's look at the denominator, (x^2 - 8). The denominator of a fraction can never be zero. Otherwise, we would have a division by zero, which is undefined in mathematics. So, we need to find the values of x that would make x^2 - 8 = 0. Solving this equation, we get x^2 = 8, which means x = sqrt(8) and x = -sqrt(8). Therefore, x cannot equal sqrt(8) or -sqrt(8). Combining both restrictions, the domain of f(x) is all real numbers greater than or equal to 2, excluding sqrt(8) and -sqrt(8). This means that if you plug in any number less than 2, the square root part will be undefined. If you plug in sqrt(8) or -sqrt(8), the denominator will be zero, causing the function to be undefined as well. Being able to identify these restrictions is fundamental when working with functions. It ensures you know which inputs will give you a valid, real output.

(d) Understanding the Implications of Undefined Values

Now, let's discuss why it's important to recognize when a function is undefined and the implications of it. Undefined values in functions often signal problems with the mathematical operations involved. When we work with real-world applications of functions, such as modeling physical phenomena, financial calculations, or scientific data, the domain restrictions become even more significant. For instance, in our function, the restriction x >= 2 could represent a physical constraint. The value within the square root could represent something like a measurement or quantity, and the domain tells us what values of that measurement make sense. If we try to use an input that leads to an undefined result, we're essentially asking a question that doesn't have a meaningful answer within the context of the problem. This can cause errors and create challenges when interpreting the results.

For example, think about plotting this function on a graph. The graph won't exist for x-values less than 2. There might also be breaks or discontinuities at x = sqrt(8) and x = -sqrt(8). This can heavily impact how you read and understand the graph. If you're using this function to model a real-world scenario, you can't just blindly plug in any number. You have to consider the domain to ensure that your model makes sense. Undefined values also highlight potential issues with the mathematical model itself. If a function is undefined for inputs that should be valid in a specific situation, then maybe the formula isn’t the best one to use for that situation. Always think about the domain and restrictions when dealing with functions. If you forget to consider these, you could end up with a nonsensical answer. Always remember the domain to avoid these traps and get to the correct answer. You can ensure that your calculations are accurate and that your conclusions are valid. The restrictions will ensure your results make sense.

In conclusion, understanding function evaluation and domain restrictions is really important. We have seen how to evaluate the functions for specific values, how to figure out when a function is defined, and how to identify the situations that make a function undefined. Keep practicing and keep asking questions, and you'll become more confident in these types of problems. You got this, guys!