Radical Function: Domain Vs. Range

Hey guys, let's dive into the wild world of functions, specifically radical ones! Today, we're tackling a question that might look a bit tricky at first glance: "Which statement best describes $f(x)=-2 \sqrt{x-7}+1$?" We're going to break down the domain and range of this function and see where the number -6 fits in. Get ready to flex those math muscles!

Understanding Domain and Range: The Core Concepts

Before we get our hands dirty with the specific function, let's make sure we're all on the same page about what domain and range actually mean. Think of the domain as all the possible input values (the 'x' values) that you can feed into a function without breaking it. It's like the allowed ingredients for your recipe – if you put in something that's not allowed, the whole thing will go kaput!

For our radical function, $f(x)=-2 \sqrt{x-7}+1$, the key player is the square root. You know how you can't take the square root of a negative number in the real number system? That's our big restriction right there. So, the expression inside the square root, which is $x-7$, has to be greater than or equal to zero. This is our first clue for finding the domain. We need $x-7 \ge 0$. If we add 7 to both sides of that inequality, we get $x \ge 7$. So, for this function, the domain includes all real numbers that are 7 or greater. Any number less than 7 will make the expression under the square root negative, and that's a no-go!

Now, let's talk about the range. The range refers to all the possible output values (the 'f(x)' or 'y' values) that the function can produce. It's like all the possible dishes you can make with your allowed ingredients. To figure out the range, we need to consider how the function transforms the input values. We start with the square root term, $\sqrt{x-7}$. Since the smallest value $x-7$ can be is 0 (when $x=7$), the smallest value $\sqrt{x-7}$ can be is also 0. The square root function itself always outputs non-negative values.

But wait, there's more! We have a coefficient of -2 multiplying the square root: $-2\sqrt{x-7}$. When you multiply a non-negative number by -2, the result will be non-positive (zero or negative). So, the term $-2\sqrt{x-7}$ can take any value that is 0 or less. The largest value this part can have is 0 (when $\sqrt{x-7}=0$).

Finally, we add 1 to the whole thing: $f(x)=-2\sqrt{x-7}+1$. Since $-2\sqrt{x-7}$ can be 0 or any negative number, adding 1 means that the smallest possible value for $f(x)$ will occur when $-2\sqrt{x-7}$ is at its most negative (which tends towards negative infinity), and the largest possible value for $f(x)$ will occur when $-2\sqrt{x-7}$ is 0. So, when $-2\sqrt{x-7}=0$, $f(x) = 0 + 1 = 1$. This means the range of our function is all real numbers that are 1 or less. We can write this as $f(x) \le 1$.

So, to recap: The domain is $x \ge 7$, and the range is $f(x) \le 1$. Keep these locked in your brain because we're going to use them to answer our main question!

Analyzing the Options: Where Does -6 Fit In?

Alright team, now we need to take our calculated domain and range and check them against the given statements about the number -6. Remember, our domain is $x \ge 7$, meaning -6 is not in the domain because it's less than 7. Our range is $f(x) \le 1$, meaning -6 is in the range because it's less than or equal to 1. Let's break down each option:

Option A: -6 is in the domain of $f(x)$ but not in the range of $f(x)$.

We just figured out that -6 is not in the domain. So, this statement is immediately incorrect. We don't even need to consider the range part.

Option B: -6 is not in the domain of $f(x)$ but is in the range of $f(x)$.

Let's check our findings. Is -6 not in the domain? Yes, because the domain requires $x \ge 7$. Is -6 in the range? Yes, because the range is $f(x) \le 1$, and -6 is indeed less than or equal to 1. This statement matches our calculations perfectly! It looks like we've found our winner, guys.

Option C: -6 is in the domain of $f(x)$ and in the range of $f(x)$.

Again, we know for a fact that -6 is not in the domain. Therefore, this statement cannot be true, even though -6 is in the range.

Option D: -6 is not in the domain of $f(x)$ and not in the range of $f(x)$.

We've established that -6 is not in the domain, which is correct. However, we also found that -6 is in the range. So, the second part of this statement is false, making the entire statement false.

Confirming the Solution and Why It Matters

So, after dissecting each option, it's crystal clear that Option B is the correct statement. To reiterate, -6 is not in the domain of $f(x)=-2 \sqrt{x-7}+1$ because the expression under the square root would become negative ($(-6)-7 = -13$), which is not allowed for real numbers. However, -6 is in the range because the function's output values are restricted to be less than or equal to 1, and -6 fits that condition.

This exercise highlights the importance of thoroughly understanding how to determine the domain and range of functions, especially those involving radicals. The domain tells you what inputs are valid, preventing mathematical impossibilities like taking the square root of a negative number. The range tells you what outputs you can expect, giving you a full picture of the function's behavior. By carefully analyzing the restrictions imposed by the square root and the effects of transformations (like the multiplication by -2 and the addition of 1), we can confidently determine the domain and range.

Remember, the domain is all about what you can put in, and the range is all about what you can get out. Keep practicing these concepts, and soon you'll be a domain and range whiz!

The Math Behind the Magic: A Deeper Dive

Let's really solidify why -6 isn't in the domain. The fundamental rule for square roots in the realm of real numbers is that the radicand (the stuff under the $\sqrt{}$ symbol) must be non-negative. For our function $f(x) = -2\sqrt{x-7} + 1$, the radicand is $x-7$. To find the domain, we set up the inequality:

$x - 7 \ge 0$

Adding 7 to both sides gives us:

$x \ge 7$

This means any value of $x$ that is less than 7 will result in a negative radicand. If we try to plug in $x = -6$, we get:

$f(-6) = -2\sqrt{-6 - 7} + 1$ $f(-6) = -2\sqrt{-13} + 1$

Since $\sqrt{-13}$ is not a real number, -6 is definitely not in the domain of $f(x)$. This confirms the first part of our correct answer.

Now, let's explore the range more rigorously. We know the square root function, $\sqrt{u}$, where $u \ge 0$, always produces a non-negative output. So, $\sqrt{x-7} \ge 0$ for all $x$ in the domain ($x \ge 7$).

When we multiply this by -2, we flip the inequality and change the sign:

$-2\sqrt{x-7} \le -2 \times 0$ $-2\sqrt{x-7} \le 0$

This tells us that the term $-2\sqrt{x-7}$ will always be zero or negative. The maximum value it can take is 0, which occurs when $x=7$.

Finally, we add 1 to this expression:

$-2\sqrt{x-7} + 1 \le 0 + 1$ $f(x) \le 1$

This confirms that the output of the function, $f(x)$, can never be greater than 1. Since the square root term can go towards infinity (as $x$ gets larger), the $-2\sqrt{x-7}$ term can go towards negative infinity. Therefore, the range encompasses all real numbers from negative infinity up to and including 1. So, $f(x) \in (-\infty, 1]$.

Since -6 is less than 1, it falls within this interval $(-\infty, 1]$. This confirms that -6 is in the range of the function.

Therefore, the statement "-6 is not in the domain of $f(x)$ but is in the range of $f(x)$" is mathematically sound and the correct description. It's all about understanding the constraints each part of the function imposes on the possible input and output values. Keep drilling these concepts, and you'll master function analysis in no time!