Function Operations: Sum, Difference, Product, And Quotient

Hey guys! Today, we're diving into the fascinating world of function operations. We're going to take a look at how to combine functions using basic arithmetic operations like addition, subtraction, multiplication, and division. Specifically, we'll be working with the functions $f(x) = \sqrt{x+1}$ and $g(x) = \sqrt{x-7}$. Our goal is to find $f + g$, $f - g$, $fg$, and $\frac{f}{g}$, and most importantly, figure out the domain for each of these new functions. Understanding the domain is crucial because it tells us where our functions are actually valid and give us real number outputs. Let's jump right in and make some mathematical magic happen!

Understanding the Basics of Function Operations

Before we start crunching numbers, let's quickly recap what function operations really mean. When we talk about adding, subtracting, multiplying, or dividing functions, we're essentially performing these operations on their outputs. For example, if we have two functions, $f(x)$ and $g(x)$:

• The sum $(f + g)(x)$ is found by adding the outputs of $f(x)$ and $g(x)$, so $(f + g)(x) = f(x) + g(x)$.
• The difference $(f - g)(x)$ is found by subtracting the outputs, giving us $(f - g)(x) = f(x) - g(x)$.
• The product $(fg)(x)$ is the result of multiplying the outputs: $(fg)(x) = f(x) imes g(x)$.
• The quotient $\left(\frac{f}{g}\right)(x)$ is obtained by dividing the outputs: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, but we need to be careful here because $g(x)$ cannot be zero!

The Importance of the Domain

Now, the domain is where things get a little more interesting. The domain of a function is the set of all possible input values (x-values) for which the function produces a real number output. When we combine functions, the domain of the resulting function is often restricted by the domains of the original functions. We have to make sure that any x-value we plug in doesn't cause any undefined operations, like dividing by zero or taking the square root of a negative number. For functions involving square roots, like the ones we're working with today, we need to ensure the expression inside the square root is non-negative (i.e., greater than or equal to zero). This is a key concept to keep in mind as we tackle our problem.

Finding $f + g$ and Its Domain

Okay, let's start with the sum of the functions, $(f + g)(x)$. Remember, we have $f(x) = \sqrt{x+1}$ and $g(x) = \sqrt{x-7}$.

To find $(f + g)(x)$, we simply add the two functions together:

$(f + g)(x) = f(x) + g(x) = \sqrt{x+1} + \sqrt{x-7}$

So, we've found the sum. Easy peasy, right? But now comes the crucial part: what's the domain of this new function? To figure this out, we need to consider the domains of both $f(x)$ and $g(x)$ separately, and then find the intersection of those domains.

Determining the Domain of $f(x)$

For $f(x) = \sqrt{x+1}$, we need to ensure that the expression inside the square root is non-negative. This means:

$x + 1 \geq 0$

Subtracting 1 from both sides gives us:

$x \geq -1$

So, the domain of $f(x)$ is all x-values greater than or equal to -1. In interval notation, this is $[-1, \infty)$.

Determining the Domain of $g(x)$

Similarly, for $g(x) = \sqrt{x-7}$, we need:

$x - 7 \geq 0$

Adding 7 to both sides, we get:

$x \geq 7$

Thus, the domain of $g(x)$ is all x-values greater than or equal to 7, which in interval notation is $[7, \infty)$.

Finding the Intersection of the Domains

Now, for $(f + g)(x)$ to be defined, both $f(x)$ and $g(x)$ must be defined. This means we need to find the intersection of the domains of $f(x)$ and $g(x)$. The domain of $f(x)$ is $[-1, \infty)$, and the domain of $g(x)$ is $[7, \infty)$. The intersection of these two intervals is the region where they overlap, which is $[7, \infty)$.

Therefore, the domain of $(f + g)(x) = \sqrt{x+1} + \sqrt{x-7}$ is $[7, \infty)$.

Finding $f - g$ and Its Domain

Next up, let's tackle the difference of the functions, $(f - g)(x)$. This is very similar to finding the sum, but instead of adding, we'll be subtracting.

$(f - g)(x) = f(x) - g(x) = \sqrt{x+1} - \sqrt{x-7}$

Again, we've found the expression for the difference. Now, what about the domain? Guess what? The domain of $(f - g)(x)$ is exactly the same as the domain of $(f + g)(x)$! Why? Because we're still dealing with the same square root functions, and we still need both $f(x)$ and $g(x)$ to be defined. So, we're still looking for the intersection of the domains of $f(x)$ and $g(x)$, which we already found to be $[7, \infty)$.

Therefore, the domain of $(f - g)(x) = \sqrt{x+1} - \sqrt{x-7}$ is also $[7, \infty)$.

Finding $fg$ and Its Domain

Time to move on to the product of the functions, $(fg)(x)$. This means we'll be multiplying $f(x)$ and $g(x)$.

$(fg)(x) = f(x) imes g(x) = \sqrt{x+1} imes \sqrt{x-7}$

We can simplify this a bit by using the property of square roots that says $\sqrt{a} imes \sqrt{b} = \sqrt{ab}$:

$(fg)(x) = \sqrt{(x+1)(x-7)} = \sqrt{x^2 - 6x - 7}$

Alright, we've got our product. Now, let's figure out the domain. Just like with the sum and difference, the domain of the product is determined by the domains of the original functions. We need both $f(x)$ and $g(x)$ to be defined, so we're back to finding the intersection of their domains. And as we already established, the intersection of the domains of $f(x)$ and $g(x)$ is $[7, \infty)$.

So, the domain of $(fg)(x) = \sqrt{x^2 - 6x - 7}$ is $[7, \infty)$.

Finding $\frac{f}{g}$ and Its Domain

Last but not least, let's tackle the quotient of the functions, $\left(\frac{f}{g}\right)(x)$. This is where things get a little trickier because we have to worry about division by zero.

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+1}}{\sqrt{x-7}}$

We've got our quotient. Now, let's dive into the domain. We know that the denominator, $g(x)$, cannot be zero. So, we need to find the values of x for which $g(x) = \sqrt{x-7} = 0$.

Accounting for Division by Zero

To find where $g(x) = 0$, we set the expression inside the square root equal to zero:

$x - 7 = 0$

Adding 7 to both sides, we get:

$x = 7$

So, $g(x) = 0$ when $x = 7$. This means we have to exclude $x = 7$ from the domain. Remember that the domain of $g(x)$ itself is $x \geq 7$. When considering the quotient $\frac{f}{g}$, we need to strictly avoid $x=7$ because it makes the denominator zero, resulting in an undefined expression.

Determining the Domain of the Quotient

We already know that the domain of $f(x)$ is $[-1, \infty)$ and the domain of $g(x)$ is $[7, \infty)$. The intersection of these domains is $[7, \infty)$. However, we also need to exclude the value $x = 7$ because it makes the denominator zero. This means the domain of $\frac{f}{g}$ will be all values greater than 7, but not including 7 itself.

In interval notation, this is $(7, \infty)$. Notice the parenthesis instead of the bracket, which indicates that 7 is not included in the domain.

Therefore, the domain of $\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x+1}}{\sqrt{x-7}}$ is $(7, \infty)$.

Wrapping It Up

Okay, guys, we've done it! We've successfully found $f + g$, $f - g$, $fg$, and $\frac{f}{g}$ for the given functions $f(x) = \sqrt{x+1}$ and $g(x) = \sqrt{x-7}$, and we've carefully determined the domain of each resulting function. Let's recap our findings:

• $(f + g)(x) = \sqrt{x+1} + \sqrt{x-7}$, Domain: $[7, \infty)$
• $(f - g)(x) = \sqrt{x+1} - \sqrt{x-7}$, Domain: $[7, \infty)$
• $(fg)(x) = \sqrt{x^2 - 6x - 7}$, Domain: $[7, \infty)$
• $\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x+1}}{\sqrt{x-7}}$, Domain: $(7, \infty)$

Remember, the key takeaway here is that understanding the domain is just as important as finding the function itself. When dealing with function operations, always consider the domains of the original functions and make sure to account for any restrictions that might arise, such as division by zero or square roots of negative numbers. You've nailed it!