Understanding Domain And Range In Mathematics

Hey guys, let's dive deep into a cool math concept: domain and range! If you're working with sets, functions, or just exploring mathematical relationships, grasping these terms is super important. Think of the domain as the input values you can feed into a mathematical expression or function, and the range as the output values you get back. It’s like a machine: you put something in (domain), and something comes out (range). Understanding these boundaries helps us predict and analyze mathematical behavior. We'll explore how to identify them, especially when dealing with inequalities and graphical representations. So, buckle up, because we're about to demystify domain and range, making those tricky math problems feel like a breeze! We'll be looking at a specific example involving a region R defined by inequalities to really nail down these concepts.

Exploring the Set R: A Visual Approach

Alright, let's get our hands dirty with a specific example to really understand how domain and range work. We're given a set $R$ defined as $R = \{(x, y): x eq 0, y eq 0 ext{ and } x+y eq 3\}$. This notation might look a bit intimidating at first, but it's actually just describing a region in the coordinate plane. The conditions $x eq 0$ and $y eq 0$ tell us we're excluding the x-axis and the y-axis. The condition $x+y eq 3$ means we're also excluding the line where $x+y$ equals 3. So, we're looking at all the points $(x, y)$ on the coordinate plane except for those on the axes and the line $x+y=3$. To really get a feel for this, imagine drawing the entire xy-plane. Now, draw the x-axis (where $y=0$) and the y-axis (where $x=0$). Erase those lines. Next, draw the line $x+y=3$. This line passes through $(3,0)$ and $(0,3)$. Now, erase that line too! What you're left with is the entire plane with these three lines removed. This is our set $R$. It's a bunch of separate regions, kind of like islands in a sea, where the sea consists of the removed lines. This visual approach is crucial because it helps us see the possible values $x$ and $y$ can take and the resulting values of $x+y$. When we talk about the domain, we're essentially asking: 'What are all the possible x-values that can exist in these remaining regions?' And for the range, we ask: 'What are all the possible y-values?' It's not just about single points; it's about the entire collection of points that satisfy the given conditions, and understanding their spread is what domain and range are all about.

Deconstructing the Inequalities: Finding the Domain

Now, let's focus on finding the domain of the set $R = \{(x, y): x eq 0, y eq 0 ext{ and } x+y eq 3\}$. The domain refers to all possible values of the first component of the ordered pair, which is $x$. So, we need to figure out what values $x$ cannot be, based on the given conditions. The first condition is $x eq 0$. This is straightforward – $x$ can be any real number except zero. The second condition is $y eq 0$. This condition doesn't directly restrict $x$, but it implies that for any given $x$, the corresponding $y$ cannot be zero. The third condition is $x+y eq 3$. This is the most interesting one for understanding restrictions on $x$. We can rewrite this as $y eq 3-x$. This means that for any given $x$, the value of $y$ cannot be equal to $3-x$. If we consider a specific value of $x$, say $x=1$, then $y$ cannot be $3-1=2$. If $x=2$, then $y$ cannot be $3-2=1$. If $x=3$, then $y$ cannot be $3-3=0$. This condition also tells us something about $x$ itself. If we were to assume that $y$ could be any real number (which it can, as long as $y eq 0$), then for a given $x$, the only time $x+y=3$ would always be true is if there was a specific value of $x$ that forced $y$ to be a particular value. However, since $y$ can be many different values (as long as $y eq 0$ and $y eq 3-x$), the restriction $x+y eq 3$ mainly affects the pairs of $(x, y)$ that are allowed. But let's think about what $x$ values are possible in our set $R$. We know $x eq 0$. What about $x+y eq 3$? Can $x$ be any other real number? Yes, it seems so. For any real number $x$ (except 0), we can always find a $y$ such that $y eq 0$ and $y eq 3-x$. For instance, if $x=5$, we need $y eq 0$ and $y eq 3-5 = -2$. We can choose $y=1$, which satisfies both. So, the conditions $x eq 0$, $y eq 0$, and $x+y eq 3$ combined mean that $x$ can be any real number except 0. Therefore, the domain of $R$ is all real numbers except 0. In interval notation, this is $(-\infty, 0) \cup (0, \infty)$. This is a key takeaway: the domain is about the permissible values of $x$.

Determining the Range of Possible Outputs

Now, let's shift our focus to the range of the set $R = \{(x, y): x eq 0, y eq 0 ext{ and } x+y eq 3\}$. The range consists of all possible values of the second component of the ordered pair, which is $y$. We need to determine what values $y$ cannot be, or rather, what values $y$ can be. The conditions are $x eq 0$, $y eq 0$, and $x+y eq 3$. The condition $y eq 0$ directly tells us that $y$ cannot be zero. What about the other conditions? The condition $x eq 0$ means that for any $y$, the corresponding $x$ cannot be zero. The condition $x+y eq 3$ can be rewritten as $x eq 3-y$. This implies that for any given $y$, the value of $x$ cannot be $3-y$. Similar to how we analyzed the domain, we need to see if there are any other restrictions on $y$. Can $y$ be any real number other than 0? Let's test a few values. Suppose we want $y=1$. We need $x eq 0$ and $x eq 3-1 = 2$. Can we find such an $x$? Yes, we can choose $x=1$. Since $x=1$ is not 0 and not 2, the point $(1,1)$ is in $R$. Now, what if we want $y=3$? We need $x eq 0$ and $x eq 3-3 = 0$. So, we need $x eq 0$. Can we find such an $x$? Yes, we can choose $x=1$. The point $(1,3)$ is in $R$. What if we want $y=-5$? We need $x eq 0$ and $x eq 3-(-5) = 8$. We can choose $x=1$. The point $(1,-5)$ is in $R$. It appears that for any real number $y$ (except 0), we can find an $x$ that satisfies the conditions. For any $y eq 0$, we need to ensure we can find an $x$ such that $x eq 0$ and $x eq 3-y$. If $y eq 3$, then $3-y eq 0$. In this case, we can choose $x$ to be any non-zero value other than $3-y$. For example, we can choose $x=1$ if $1 eq 3-y$. If $y=3$, then $3-y=0$. We need $x eq 0$. So for $y=3$, we can pick any $x eq 0$, like $x=1$. The point $(1,3)$ is in $R$. So, it seems $y$ can be any real number except 0. Therefore, the range of $R$ is all real numbers except 0. In interval notation, this is $(-\infty, 0) \cup (0, \infty)$. The range is concerned with all the possible output values $y$ can take.

Analyzing the Original Problem: A Crucial Distinction

Let's go back to the original problem description and carefully examine the set $R$: $R = {(x, y): x \geq 0, y

\geq 0 ext{ and } x+y

\leq 3}$. This is a *different* set than the one we just analyzed! This is a super common pitfall in math – the slightest change in symbols can drastically alter the solution. Here, instead of strict inequalities ($ eq $), we have non-strict inequalities ($

\geq $ and $

\leq $). This means we are including the boundaries. Let's break down these new conditions:

• **$x

\geq 0$**: This means $x$ can be zero or any positive number. Geometrically, this restricts our points to the right half of the coordinate plane, including the y-axis.

• **$y

\geq 0$**: This means $y$ can be zero or any positive number. Geometrically, this restricts our points to the upper half of the coordinate plane, including the x-axis.

When we combine $x

\geq 0$ and $y

\geq 0$, we are looking at the first quadrant of the coordinate plane, including the non-negative parts of the x and y axes.

• **$x+y

\leq 3$**: This means the sum of $x$ and $y$ can be 3, or it can be any value less than 3. Geometrically, this describes the region on and below the line $x+y=3$. This line passes through the points $(3,0)$ and $(0,3)$.

So, our set $R$ is the region in the first quadrant that lies on or below the line $x+y=3$. This forms a triangle with vertices at $(0,0)$, $(3,0)$, and $(0,3)$. It includes the edges and the interior of this triangle. This is a much more confined region than the previous one!

Finding the Domain for the New Set R

Now, let's find the domain for this new set $R = {(x, y): x

\geq 0, y

\geq 0 ext{ and } x+y

\leq 3}$. The domain consists of all possible $x$-values within this triangular region. Looking at our triangle with vertices $(0,0)$, $(3,0)$, and $(0,3)$:

• The smallest $x$-value in this region is $0$ (along the y-axis).
• The largest $x$-value in this region is $3$ (at the point $(3,0)$).

Since the region is a solid triangle (including its interior), $x$ can take on any value between 0 and 3, inclusive. For any $x$ between 0 and 3, we can find a corresponding $y$ such that $y

\geq 0$ and $x+y

\leq 3$. For example, if $x=1$, then $y$ can range from $0$ up to $3-1=2$. So, points like $(1,0)$, $(1,1)$, and $(1,2)$ are all in our set $R$. This confirms that $x=1$ is a valid domain value.

Therefore, the domain of $R$ is all $x$ such that $0

\leq x

\leq 3$. In interval notation, this is $[0,3]$. This matches option C!

Determining the Range for the New Set R

Finally, let's determine the range for this triangular region $R = {(x, y): x

\geq 0, y

\geq 0 ext{ and } x+y

\leq 3}$. The range consists of all possible $y$-values within this region.

• The smallest $y$-value in this region is $0$ (along the x-axis).
• The largest $y$-value in this region is $3$ (at the point $(0,3)$).

Similar to the domain, since the region is a solid triangle, $y$ can take on any value between 0 and 3, inclusive. For any $y$ between 0 and 3, we can find a corresponding $x$ such that $x

\geq 0$ and $x+y

\leq 3$. For instance, if $y=1$, then $x$ can range from $0$ up to $3-1=2$. So, points like $(0,1)$, $(1,1)$, and $(2,1)$ are all in our set $R$. This confirms that $y=1$ is a valid range value.

Therefore, the range of $R$ is all $y$ such that $0

\leq y

\leq 3$. In interval notation, this is $[0,3]$.

Conclusion: Putting It All Together

So, guys, we've thoroughly analyzed the set $R = {(x, y): x

\geq 0, y

\geq 0 ext{ and } x+y

\leq 3}$. We found that the domain is all possible $x$-values, which are $x

\in [0,3]$. We also found that the range is all possible $y$-values, which are $y

\in [0,3]$.

Let's look back at the options provided:

A. Domain $=(0,3)$ B. Range $=(-\infty, 3]$ C. Domain $=[0,3]$ D. Range $=(3,

\infty)$

Our calculations clearly show that the Domain is $[0,3]$. This matches option C. Options A, B, and D are incorrect because they describe different sets of values or use incorrect interval notation for this specific region.

Understanding domain and range is fundamental in mathematics, helping us define the boundaries of our variables and the possible outcomes of our functions or sets. Keep practicing, and you'll become a pro at spotting them in no time!