Finding The Domain And Range Of F(x) = \sqrt{ \frac{x - 3}{x - 5} }

Hey guys, I'm super stuck on this math problem and could really use some help. I'm trying to figure out the domain and range of the function $\bf{f(x) = \sqrt{ \frac{x - 3}{x - 5} }}$ and my brain is just spinning! It seems like a complex function and I'm not sure where to even start. If anyone has any experience figuring out these function properties, your insights would be a lifesaver! So, I'm putting out an SOS to the math whizzes out there – can someone break down how to solve this step by step? I'm really hoping to understand the process so I can tackle similar problems on my own in the future. I've watched a few videos, but I think I need a more detailed explanation, maybe with someone walking me through the logic. I am trying to nail down the domain and range function, but this example makes me dizzy. Could you simplify the steps in determining the domain, explaining why we need to consider certain restrictions and how they arise in this specific function. Maybe you can clarify how the square root and the fraction within it influence the possible input values, and also, I am a bit confused on figuring out the range, especially with the combination of the square root and fraction. Are there specific techniques or considerations that apply when finding the range of such functions? How do you tackle the interplay between the function's components and the resulting output values? Any clear explanations or examples would be greatly appreciated. I’m open to any tips, tricks, or clear explanations you can throw my way! Let's solve this together!

Understanding the Domain of f(x) = \sqrt{ \frac{x - 3}{x - 5} }

Okay, let's dive into finding the domain of the function $\bf{f(x) = \sqrt{ \frac{x - 3}{x - 5} }}$. The domain, as you guys probably know, is basically the set of all possible 'x' values that we can plug into the function without breaking any math rules. For this specific function, we have two major rules we need to keep in mind:

1. We can't take the square root of a negative number. Remember, in the real number system, the square root of a negative number is undefined. That means the expression inside the square root, in this case, $\frac{x - 3}{x - 5}$, must be greater than or equal to zero. So, we have the inequality: $\frac{x - 3}{x - 5} \ge 0$.
2. We can't divide by zero. Denominators can't be zero! If you try to divide a number by zero, you get an undefined result. In our function, the denominator is $(x - 5)$. So, $x - 5$ cannot equal zero, which means $x \ne 5$.

So, to find the domain, we need to solve the inequality $\frac{x - 3}{x - 5} \ge 0$ while keeping in mind that $x \ne 5$. To solve this inequality, we can use a sign chart. First, we find the critical points, which are the values of x that make the numerator or denominator equal to zero. These are $x = 3$ and $x = 5$.

Now, we create a number line and mark these critical points. This divides the number line into three intervals: $(-\infty, 3)$, $(3, 5)$, and $(5, \infty)$. We need to test a value from each interval in the inequality $\frac{x - 3}{x - 5} \ge 0$ to see if the inequality holds true.

• Interval $(-\infty, 3)$: Let's pick $x = 0$. Then, $\frac{0 - 3}{0 - 5} = \frac{-3}{-5} = \frac{3}{5}$, which is greater than zero. So, this interval is part of the domain.
• Interval $(3, 5)$: Let's pick $x = 4$. Then, $\frac{4 - 3}{4 - 5} = \frac{1}{-1} = -1$, which is less than zero. So, this interval is NOT part of the domain.
• Interval $(5, \infty)$: Let's pick $x = 6$. Then, $\frac{6 - 3}{6 - 5} = \frac{3}{1} = 3$, which is greater than zero. So, this interval is part of the domain.

Since the inequality is greater than or equal to zero, we include the value $x = 3$ in the domain (because the numerator can be zero). However, we exclude $x = 5$ because it makes the denominator zero. Therefore, the domain of the function is $(-\infty, 3] \cup (5, \infty)$. Basically, any number less than or equal to 3, or any number greater than 5, can be plugged into the function without causing any math problems.

Cracking the Range of f(x) = \sqrt{ \frac{x - 3}{x - 5} }

Alright, now let's tackle the range of the function $\bf{f(x) = \sqrt{ \frac{x - 3}{x - 5} }}$. The range, in simple terms, is the set of all possible output values (y-values) that the function can produce. This can be a bit trickier to figure out than the domain, but we can definitely do it! With functions involving square roots and fractions, we need to think carefully about the possible outputs.

First, let's consider the square root. The square root function itself only produces non-negative values. This is super important! The output of a square root is always zero or a positive number. So, we know right away that the range of our function will be limited to values greater than or equal to zero. In mathematical notation, we know the range will be something like $[0, \ldots)$. The question is, what's the upper limit?

To figure out the upper limit, we need to analyze the expression inside the square root, which is $\frac{x - 3}{x - 5}$. Let's call this expression $g(x) = \frac{x - 3}{x - 5}$. We want to see what values $g(x)$ can take on within the domain we already found, which is $(-\infty, 3] \cup (5, \infty)$.

It's helpful to rewrite $g(x)$ to make its behavior clearer. We can do this using a little algebraic trick. We can rewrite the numerator as $(x - 5) + 2$. Then, we have:

$\bf{g(x) = \frac{x - 3}{x - 5} = \frac{(x - 5) + 2}{x - 5} = \frac{x - 5}{x - 5} + \frac{2}{x - 5} = 1 + \frac{2}{x - 5}}$

Now, this form is much easier to analyze! Notice that $g(x)$ is equal to 1 plus the fraction $\frac{2}{x - 5}$. Let's think about what happens to this fraction as x varies within our domain.

• As x approaches 5 from the right (i.e., x is a bit bigger than 5): The denominator $(x - 5)$ becomes a small positive number. So, $\frac{2}{x - 5}$ becomes a large positive number, and $g(x)$ becomes a large number greater than 1.
• As x approaches infinity: The fraction $\frac{2}{x - 5}$ approaches zero, so $g(x)$ approaches 1.
• As x approaches 3 from the left (i.e., x is a bit smaller than 3): The denominator $(x - 5)$ is negative. The smallest value of x in this domain is 3, which makes the numerator 0. When x = 3, g(3) = (3-3)/(3-5) = 0/-2 = 0. Since the value inside the square root can be 0, then f(x) can be 0.
• As x approaches negative infinity: The fraction $\frac{2}{x - 5}$ approaches zero, so $g(x)$ approaches 1.

So, $g(x)$ can take on values from 0 (when x = 3) and approaches 1 as x tends to $ \pm \infty $, but can also be bigger than 1 as x approaches 5 from the right.

Now, remember that $f(x) = \sqrt{g(x)}$. Since $g(x)$ can be 0, $f(x)$ can be $\sqrt{0} = 0$. As $g(x)$ becomes a large positive number, $f(x)$ also becomes a large positive number (though growing more slowly because of the square root). As $g(x)$ approaches 1, $f(x)$ approaches $\sqrt{1} = 1$. Therefore, the range of the function is $[0, 1)$.

Final Answer

So, to wrap it all up, we've found that for the function $\bf{f(x) = \sqrt{ \frac{x - 3}{x - 5} }}$:

• Domain: $(-\infty, 3] \cup (5, \infty)$
• Range: $[0, 1)$

I hope this explanation helps you understand how to find the domain and range of this function! Remember, the key is to break down the problem into smaller parts, think about the restrictions imposed by square roots and fractions, and analyze the behavior of the function as x varies. You got this!