Function Operations: Sum, Difference, Product, And Quotient

Hey math enthusiasts! Let's dive into the awesome world of function operations. We're going to explore how to combine functions using addition, subtraction, multiplication, and division. Plus, we'll figure out the domain for each new function we create. This is super important stuff, so let's get started, shall we?

We'll be working with two functions: $f(x) = \sqrt{x+6}$ and $g(x) = \sqrt{x+5}$. Our goal is to find $f+g$, $f-g$, $fg$, and $\frac{f}{g}$. The domain, remember, is just the set of all possible input values (x-values) for which the function is defined. For square root functions, like the ones we have, the expression inside the square root (the radicand) must be greater than or equal to zero. So, let's break it down step by step and make this super easy to understand.

Finding $f+g$ and its Domain

First up, let's find $f+g$. This means we're adding the two functions together. It's as simple as it sounds: $(f+g)(x) = f(x) + g(x)$. So, plugging in our functions, we get:

$(f+g)(x) = \sqrt{x+6} + \sqrt{x+5}$.

Now, about the domain. We need to consider the domains of both $f(x)$ and $g(x)$ separately, and then find the intersection. Remember, for the square root to be real, the stuff inside the square root must be non-negative (greater than or equal to zero). For $f(x) = \sqrt{x+6}$, we need $x+6 \ge 0$, which means $x \ge -6$. For $g(x) = \sqrt{x+5}$, we need $x+5 \ge 0$, meaning $x \ge -5$. So, for $(f+g)(x)$, both square roots must be defined. Therefore, we need to satisfy both $x \ge -6$ and $x \ge -5$. Because $-5$ is greater than $-6$, the stricter condition is $x \ge -5$. This is because if $x$ is greater or equal to $-5$, it will also be greater or equal to $-6$. Therefore, the domain of $(f+g)(x)$ is all $x$ such that $x \ge -5$, or in interval notation, $[-5, \infty)$.

To really get this, imagine plugging in a number. If you choose -6, then $\sqrt{x+6}$ is 0 but $\sqrt{x+5}$ is $\sqrt{-1}$, which is not a real number. If you choose -5, $\sqrt{x+5}$ becomes zero, and $\sqrt{x+6}$ equals one. So, $-5$ is perfectly fine. The domain ensures that our function behaves in the realm of real numbers and we avoid those pesky imaginary numbers.

Finding $f-g$ and its Domain

Next, let's tackle $f-g$. This is the subtraction of the functions: $(f-g)(x) = f(x) - g(x)$. So:

$(f-g)(x) = \sqrt{x+6} - \sqrt{x+5}$.

For the domain of $(f-g)(x)$, the logic is the same as with $f+g$. We still need both square roots to be defined. So, again, we require $x+6 \ge 0$ and $x+5 \ge 0$. This leads us back to $x \ge -6$ and $x \ge -5$, respectively. Because we want both square roots to be real, we must satisfy both conditions. As before, $x \ge -5$ is the more restrictive of the two, meaning the domain of $(f-g)(x)$ is $x \ge -5$, or in interval notation, $[-5, \infty)$.

It's like this: you need both functions to 'work' for the subtraction to make sense. If either square root becomes imaginary, the whole thing falls apart in the real number system. Keep in mind that the minus sign in the middle changes the result of the function, but it doesn't change the domain. The domain is determined by where the individual parts of the function are defined.

Finding $f g$ and its Domain

Time to multiply! Now we'll find $fg$, which means $(fg)(x) = f(x) \cdot g(x)$. So:

$(fg)(x) = \sqrt{x+6} \cdot \sqrt{x+5}$.

This can also be written as $(fg)(x) = \sqrt{(x+6)(x+5)}$.

For the domain of the product, the rule remains consistent. Both square roots must be defined. So, we still need $x \ge -6$ and $x \ge -5$. The more restrictive condition dictates our domain. Therefore, the domain of $(fg)(x)$ is $x \ge -5$, or in interval notation, $[-5, \infty)$. The multiplication doesn't change the fundamental requirements for the inputs. Each square root needs to have a non-negative radicand to be defined, and that's the bottom line.

Think of it as creating a new function that builds on the old ones. The new function can only do what the original functions could do. If $f(x)$ is undefined, then the product function is also undefined, no matter what $g(x)$ is doing.

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

Finally, let's find the quotient $\frac{f}{g}$, which is $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$. So:

$(\frac{f}{g})(x) = \frac{\sqrt{x+6}}{\sqrt{x+5}}$.

Things get a little trickier here! For the domain, we still need $x+6 \ge 0$, or $x \ge -6$, to keep the numerator defined. And we still need $x+5 \ge 0$, or $x \ge -5$, to keep the denominator defined. However, there's a new consideration: we can't divide by zero! Therefore, we must ensure that the denominator, $\sqrt{x+5}$, is not equal to zero. This means $x+5 > 0$, and so $x > -5$. We've essentially removed the value of -5 from our domain, because it makes the denominator equal to zero. If x is equal to $-5$, the expression becomes $\frac{\sqrt{-5+6}}{\sqrt{-5+5}} = \frac{\sqrt{1}}{0}$, which is undefined.

So, for the quotient, we need $x \ge -6$ and $x > -5$. Because $x$ must be strictly greater than -5, this condition overrides $x \ge -6$. So, the domain of $(\frac{f}{g})(x)$ is all $x$ such that $x > -5$, or in interval notation, $(-5, \infty)$. Note the parenthesis instead of a bracket: we exclude -5.

Dividing adds an extra layer of complexity because of that zero in the denominator issue. Make sure that when you're dealing with fractions, you always check for values that might make the denominator zero. Those values are off-limits for the domain.

Summary of Results

Let's recap what we've found:

• $(f+g)(x) = \sqrt{x+6} + \sqrt{x+5}$, Domain: $[-5, \infty)$
• $(f-g)(x) = \sqrt{x+6} - \sqrt{x+5}$, Domain: $[-5, \infty)$
• $(fg)(x) = \sqrt{x+6} \cdot \sqrt{x+5}$, Domain: $[-5, \infty)$
• $(\frac{f}{g})(x) = \frac{\sqrt{x+6}}{\sqrt{x+5}}$, Domain: $(-5, \infty)$

Awesome work, everyone! You've successfully navigated the world of function operations and domains. Keep practicing, and you'll become a function guru in no time. If you have any questions, feel free to ask, and happy calculating!