Range Of Y = Cbrt(x+8)

Hey everyone! Today, we're diving deep into a super common math question: What is the range of the function $y=\sqrt[3]{x+8}$? This might sound a little intimidating with that cube root and plus eight, but trust me, guys, once you break it down, it's actually pretty straightforward. We'll be looking at the options provided: A. $-8$, B. $2 \leq y<\infty$, C. $0 \leq y<\infty$, and D. $-\infty$. Our goal is to figure out which of these best describes all the possible y-values that our function can produce. We're going to tackle this step-by-step, so even if you're not a math whiz, you'll be able to follow along and understand the logic behind it. We'll explore what a function's range actually means, how cube roots behave, and how that '+8' inside the root affects things. By the end of this, you'll be able to confidently determine the range of similar functions. So, grab your favorite beverage, get comfortable, and let's get this math party started!

Understanding Function Range

Alright guys, before we jump into the nitty-gritty of our specific function, let's get crystal clear on what the range of a function actually is. Think of a function like a machine. You put something in (that's the input, or x-value), and the machine does its thing and spits something out (that's the output, or y-value). The domain of a function is the set of all possible inputs (all the x-values you're allowed to put into the machine). The range, on the other hand, is the set of all possible outputs (all the y-values that the machine can spit out). It's like asking, "What are all the things this machine can possibly make?" For our function, $y=\sqrt[3]{x+8}$, we're interested in all the possible y-values we can get when we plug in different x-values. We need to consider what kind of numbers the cube root can produce and how the addition of 8 might shift things. It's crucial to remember that for real-valued functions (which is what we're dealing with here), we're only interested in real number outputs. We don't want any imaginary numbers popping up unless the problem specifically asks for them, which this one doesn't.

The Behavior of Cube Roots

Now, let's talk about the star of the show: the cube root function, $\sqrt[3]{x}$. This is a really important piece of the puzzle for understanding the range of $y=\sqrt[3]{x+8}$. Unlike square roots, which have restrictions (you can't take the square root of a negative number and get a real result), cube roots are way more flexible. You can take the cube root of any real number, positive or negative, and get a real number back. For example, $\sqrt[3]{8} = 2$ because $2 \times 2 \times 2 = 8$. But also, $\sqrt[3]{-8} = -2$ because $(-2) \times (-2) \times (-2) = -8$. This is a huge difference from square roots. What does this mean for the range of just $y=\sqrt[3]{x}$? Well, as you input larger and larger positive numbers for $x$, the cube root gets larger and larger, heading towards positive infinity ($+\infty$). Think about $\sqrt[3]{1000} = 10$, $\sqrt[3]{1,000,000} = 100$. You get the idea. Similarly, as you input larger and larger negative numbers for $x$, the cube root gets larger and larger in the negative direction, heading towards negative infinity ($-\infty$). For instance, $\sqrt[3]{-1000} = -10$, $\sqrt[3]{-1,000,000} = -100$. So, for the basic cube root function $y=\sqrt[3]{x}$, the range is all real numbers. This means the y-values can be anything from $-\infty$ to $+\infty$. This is our baseline understanding, and now we need to see how the '+8' inside the function affects this.

Analyzing $y=\sqrt[3]{x+8}$

Okay, guys, we've established that the basic cube root function, $y=\sqrt[3]{x}$, has a range of all real numbers ($-\infty$ to $+\infty$). Now, let's bring in our specific function: $y=\sqrt[3]{x+8}$. What difference does that '+8' inside the cube root make? When we're looking at the range (the possible y-values), we're asking what outputs this function can produce. Think about the expression inside the cube root: $x+8$. Since $x$ can be any real number, $x+8$ can also be any real number. For example, if $x=0$, $x+8=8$, and $y=\sqrt[3]{8}=2$. If $x=-8$, $x+8=0$, and $y=\sqrt[3]{0}=0$. If $x=-16$, $x+8=-8$, and $y=\sqrt[3]{-8}=-2$. Because the expression $x+8$ can produce any real number as its value (just like $x$ alone could), and because the cube root function can handle any real number as input, the output of the cube root, $y$, can also be any real number. The '+8' inside the cube root actually affects the domain in a different way if it were under a square root (e.g., $\sqrt{x+8}$ requires $x+8 \geq 0$, so $x \geq -8$). However, for a cube root, there's no such restriction on the input. The '+8' causes a horizontal shift of the graph of $y=\sqrt[3]{x}$ eight units to the left. But a horizontal shift does not change the set of possible y-values the function can output. Since the basic cube root function's range is all real numbers, and the '+8' only shifts the graph left or right, the range of $y=\sqrt[3]{x+8}$ remains all real numbers. So, the y-values can extend from negative infinity to positive infinity.

Evaluating the Options

Let's look back at the options provided for the range of $y=\sqrt[3]{x+8}$: A. $-8$, B. $2 \leq y<\infty$, C. $0 \leq y<\infty$, D. $-\infty$. We've reasoned that the range of this function is all real numbers, which means $y$ can be any value from $-\infty$ to $+\infty$. Let's examine each option:

• A. $-8$: This is just a single number. The range is a set of numbers, not a single value. So, this is incorrect.
• B. $2 \leq y<\infty$: This option suggests that the y-values start at 2 and go up to infinity. This would be true if we were dealing with something like $y = \sqrt{x+8}$ and considering only $x \geq -8$, which would give $y \geq 0$, or if we had a transformation that shifted the graph upwards. But our cube root function is not restricted this way.
• C. $0 \leq y<\infty$: This option suggests that the y-values start at 0 and go up to infinity. This is also too restrictive. We know from our earlier discussion that cube roots can produce negative numbers (e.g., if $x+8 = -27$, then $y=-3$). So, the range cannot start at 0.
• D. $-\infty$: This option seems to be incomplete or misunderstood. While it mentions infinity, it only includes the negative side and doesn't represent the full spectrum of possible y-values. However, if we interpret this as representing the set of all real numbers, which is $(-\infty, \infty)$, then it's the closest in concept to what we've deduced. Often, in multiple-choice questions, one option might be presented in a way that needs careful interpretation. The true range is all real numbers. Let's re-evaluate how the options are presented. If Option D is meant to imply $(-\infty, \infty)$, it would be the correct choice. If it literally means only negative infinity, it's incorrect. Let's assume there might be a typo or a standard way these are presented. A common way to express