Are F(x) And G(x) Inverse Functions? Let's Find Out!

Hey math enthusiasts! Ever wondered if two functions are like best buddies, perfectly undoing each other? That's what we're diving into today with our trusty functions, $f(x) = 6x + 3$ and $g(x) = \frac{1}{6}(x - 3)$. We're going to figure out if these two are indeed inverse functions. It's a super cool concept in mathematics, and understanding it can unlock a whole new level of problem-solving.

So, what exactly does it mean for two functions to be inverses? Think of it like this: if a function $f$ takes an input and gives you an output, its inverse function, let's call it $f^{-1}$, takes that output and gives you back the original input. They're like a lock and its key! If $f(a) = b$, then $f^{-1}(b) = a$. We'll be testing this relationship for our given $f(x)$ and $g(x)$.

Part (a): Finding $f(g(x))$

Alright guys, let's get our hands dirty and compute $f(g(x))$. This is what we call a composite function. It means we're going to take the entire function $g(x)$ and plug it into the $x$ of function $f(x)$. Sounds a bit mind-bendy, but it's straightforward once you get the hang of it. Our $f(x)$ is $6x + 3$, and our $g(x)$ is $\frac{1}{6}(x - 3)$.

So, to find $f(g(x))$, we replace every 'x' in $f(x)$ with the expression for $g(x)$.

$f(g(x)) = 6\left(g(x)\right) + 3$

Now, let's substitute $g(x) = \frac{1}{6}(x - 3)$ into this equation:

$f(g(x)) = 6\left(\frac{1}{6}(x - 3)\right) + 3$

See what happened there? We replaced the $x$ inside $f(x)$ with the whole $\frac{1}{6}(x - 3)$ thing. Now, we simplify. The '6' outside the parenthesis multiplies the $\frac{1}{6}$ inside. These two '6's are going to cancel each other out, which is pretty neat!

$f(g(x)) = \cancel{6} \times \frac{1}{\cancel{6}}(x - 3) + 3$

This leaves us with:

$f(g(x)) = (x - 3) + 3$

And simplifying further:

$f(g(x)) = x - 3 + 3$

$f(g(x)) = x$

Wowza! So, $f(g(x))$ simplifies to just $x$. This is a huge clue, guys. When the composite function $f(g(x))$ simplifies to $x$, it's a strong indicator that $g(x)$ might be the inverse of $f(x)$. But we're not done yet; we need to check the other way around too!

Part (b): Finding $g(f(x))$

Now, we do the same thing but in reverse. We're going to find $g(f(x))$. This means we take the entire function $f(x)$ and plug it into the $x$ of function $g(x)$. Remember, the order matters in composite functions, so $f(g(x))$ and $g(f(x))$ don't always give the same result. But if they are inverses, they will both simplify to $x$.

Our $g(x)$ is $\frac{1}{6}(x - 3)$, and our $f(x)$ is $6x + 3$. To find $g(f(x))$, we replace every 'x' in $g(x)$ with the expression for $f(x)$.

$g(f(x)) = \frac{1}{6}\left(f(x) - 3\right)$

Let's substitute $f(x) = 6x + 3$ into this equation:

$g(f(x)) = \frac{1}{6}\left((6x + 3) - 3\right)$

Notice how we put $6x + 3$ inside the parentheses where the $x$ used to be in $g(x)$. Now, let's simplify the expression inside the parentheses first. We have $+3$ and $-3$, which cancel each other out.

$g(f(x)) = \frac{1}{6}\left(6x + \cancel{3} - \cancel{3}\right)$

$g(f(x)) = \frac{1}{6}(6x)$

Now, we have $\frac{1}{6}$ multiplying $6x$. Just like before, these '6's are going to do some canceling!

$g(f(x)) = \frac{1}{\cancel{6}} \times \cancel{6}x$

And that simplifies to:

$g(f(x)) = x$

Boom! Just like that, $g(f(x))$ also simplifies to $x$. This is exactly what we were hoping for, guys! When both $f(g(x))$ and $g(f(x))$ simplify to $x$, it means our functions $f$ and $g$ are indeed inverses of each other.

Part (c): Determining if $f$ and $g$ are Inverses

So, after all that calculation, what's the verdict? Are $f(x) = 6x + 3$ and $g(x) = \frac{1}{6}(x - 3)$ inverse functions? Based on our findings from parts (a) and (b), the answer is a resounding YES!

In mathematics, two functions, say $f$ and $g$, are considered inverses of each other if and only if $f(g(x)) = x$ for all $x$ in the domain of $g$, AND $g(f(x)) = x$ for all $x$ in the domain of $f$. We've just proven both of these conditions to be true for our functions.

Let's quickly recap why this works. Function $f(x) = 6x + 3$ takes an input, multiplies it by 6, and then adds 3. Its inverse, $g(x) = \frac{1}{6}(x - 3)$, does the exact opposite operations in the reverse order. It first subtracts 3 from the input, and then divides the result by 6 (which is the same as multiplying by $\frac{1}{6}$).

Think about it: if you start with a number, say 5:

1. Apply $f$: $f(5) = 6(5) + 3 = 30 + 3 = 33$.
2. Now, apply $g$ to the result (33): $g(33) = \frac{1}{6}(33 - 3) = \frac{1}{6}(30) = 5$.

You got your original number back! This is the essence of inverse functions. They 'undo' each other.

This concept is super useful in various areas of math and science. For instance, if you have a function that models a physical process, its inverse function can help you figure out the initial conditions or parameters that led to a certain outcome. It's like having a rewind button for your equations!

So, to sum it all up, we started with $f(x)=6x+3$ and $g(x)=\frac{1}{6}(x-3)$. We computed $f(g(x))$ and found it equals $x$. Then, we computed $g(f(x))$ and also found it equals $x$. Because both composite functions resulted in $x$, we can confidently conclude that $f$ and $g$ are indeed inverse functions. Pretty cool, right? Keep practicing these, and you'll become a master of inverses in no time! Happy calculating, everyone!