Functions With Single Number Range

Hey guys! Let's dive into a fascinating algebra problem today. We're going to explore functions and figure out which one has a range that's just a single number. This might sound a bit abstract, but trust me, it’s super cool when you see how it works. We’ll break down each option step by step, so you can totally nail this kind of question in the future. Let’s get started!

Understanding the Range of a Function

Before we jump into the specific functions, let's quickly recap what the range of a function actually means. Simply put, the range is the set of all possible output values (y-values) that a function can produce. Think of it like this: you put in some x-values, the function does its magic, and the range is all the possible results you can get out. For instance, if a function always gives you a result between 0 and 1, then its range is the interval [0, 1].

So, when we’re asked which function has a range consisting of a single number, we’re looking for a function that always spits out the same y-value, no matter what x-value we plug in. This means the function's output doesn't change; it's constant. Functions like these are pretty special, and they often have unique properties that make them stand out. Understanding ranges is crucial in algebra because it helps us grasp how functions behave and what kind of results to expect. It's not just about plugging in numbers; it's about seeing the bigger picture of what the function is doing. Okay, with this in mind, let's examine the given options and see which one fits the bill. We'll go through each function, analyze its behavior, and determine its range. By the end of this, you'll be a range-finding pro!

Analyzing Option A: $y = \\sqrt[4]{-x}$

Let's start with option A: $y = \sqrt[4]{-x}$. This function involves a fourth root, which is a type of radical. Now, when dealing with even roots (like square roots, fourth roots, etc.), we need to be careful about the numbers we put inside. Remember, we can’t take an even root of a negative number and get a real number result. So, what does this mean for our function? Well, the expression inside the fourth root, which is $-x$, must be non-negative (i.e., greater than or equal to zero). Mathematically, we write this as $-x \\geq 0$. If we solve this inequality, we find that $x \\leq 0$. This tells us that the domain of the function (the set of possible x-values) consists of all non-positive numbers. We can only plug in zero or negative numbers into this function.

Now, let's think about the possible output values (the range). When $x = 0$, we have $y = \sqrt[4]{-0} = \sqrt[4]{0} = 0$. That’s one value we know is in the range. What happens when we plug in other negative numbers? For example, if $x = -1$, we get $y = \sqrt[4]{-(-1)} = \sqrt[4]{1} = 1$. If $x = -16$, we get $y = \sqrt[4]{-(-16)} = \sqrt[4]{16} = 2$. Notice a pattern? As we plug in more negative values for x, the output y-values increase. The smallest possible output is 0, and the outputs can grow indefinitely as x becomes more negative. Therefore, the range of this function is all non-negative numbers, or $[0, \\infty)$. This is a whole range of numbers, not just a single one, so option A is not our answer.

Analyzing Option B: $y = \\sqrt[3]{-x}$

Next up, let's consider option B: $y = \sqrt[3]{-x}$. This function involves a cube root, which is an odd root. Unlike even roots, odd roots can handle negative numbers inside without any problems. This is because a negative number multiplied by itself three times (or any odd number of times) will still be negative. So, what does this mean for the domain of our function? Well, since we can take the cube root of any real number, the domain of this function is all real numbers. We can plug in any value for x, whether it’s positive, negative, or zero.

Now, let’s think about the range. When $x = 0$, we have $y = \sqrt[3]{-0} = \sqrt[3]{0} = 0$. If $x = 1$, we get $y = \sqrt[3]{-1} = -1$. And if $x = -1$, we get $y = \sqrt[3]{-(-1)} = \sqrt[3]{1} = 1$. Notice that we can get both positive and negative output values. As x becomes very large, the output becomes a large negative number, and as x becomes a large negative number, the output becomes a large positive number. This means the range of the function includes all real numbers, from negative infinity to positive infinity $(-\\infty, \\infty)$. Again, this is a whole range of numbers, not just a single one, so option B is not our answer either.

Analyzing Option C: $y = \\sqrt[4]{-x^2}$

Let’s move on to option C: $y = \sqrt[4]{-x^2}$. This one is interesting because it combines a fourth root (like in option A) with a squared term. Remember, we have to be extra careful with even roots. The expression inside the fourth root, which is $-x^2$, must be non-negative. So, we need $-x^2 \\geq 0$. Now, think about what happens when you square a number. Squaring any real number (positive, negative, or zero) always gives you a non-negative result. That means $x^2$ is always greater than or equal to zero. But we have a negative sign in front of it, so $-x^2$ is always less than or equal to zero. For $-x^2$ to be non-negative (as required by the fourth root), it must be equal to zero. This happens only when $x = 0$.

So, the domain of this function is just a single value: $x = 0$. There’s only one number we can plug in. What happens when we plug it in? We get $y = \sqrt[4]{-0^2} = \sqrt[4]{0} = 0$. That’s it! The only possible output value is 0. Therefore, the range of this function consists of just a single number: 0. We’ve found our answer! Option C has a range consisting of only one number. But, just to be thorough, let's quickly look at option D as well.

Analyzing Option D: $y = \\sqrt[3]{-x^2}$

Finally, let’s examine option D: $y = \sqrt[3]{-x^2}$. This function involves a cube root and a squared term, similar to option C, but with a cube root instead of a fourth root. As we discussed earlier, cube roots can handle negative numbers without any issues. However, we still have that squared term inside. The expression inside the cube root is $-x^2$. We know that $x^2$ is always non-negative, so $-x^2$ is always non-positive (less than or equal to zero).

This means that the values inside the cube root will always be zero or negative. When $x = 0$, we have $y = \sqrt[3]{-0^2} = \sqrt[3]{0} = 0$. When $x$ is any other real number, $x^2$ will be positive, so $-x^2$ will be negative. For example, if $x = 1$, we get $y = \sqrt[3]{-1^2} = \sqrt[3]{-1} = -1$. If $x = -1$, we also get $y = \sqrt[3]{-(-1)^2} = \sqrt[3]{-1} = -1$. The outputs will always be zero or negative. Therefore, the range of this function includes 0 and all negative numbers, which can be written as $(-\\infty, 0]$. This is a range of numbers, not a single number, so option D is not our answer.

Conclusion: The Function with a Single-Number Range

Alright guys, we've gone through all the options, and it's clear that the function with a range consisting of only one number is option C: $y = \sqrt[4]{-x^2}$. This function is special because the combination of the fourth root and the squared term restricts the domain to a single value ($x = 0$), which in turn results in a single output value ($y = 0$). Understanding the domains and ranges of functions is a key skill in algebra, and you've just leveled up your knowledge! Keep practicing, and you'll become a function master in no time.

Which of the following functions has a range consisting of a single number?

A) $y = \sqrt[4]{-x}$; Б) $y = \sqrt[3]{-x}$; В) $y = \sqrt[4]{-x^2}$; Г) $y = \sqrt[3]{-x^2}$