- anonymous

Domain Values question. (Picture of question in thread)

- schrodinger

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- anonymous

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- anonymous

- jtvatsim

K, let's see...

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## More answers

- jtvatsim

First of all, we should write down what the composition looks like, it's a bit ugly, but have you done that step?

- anonymous

No, I didn't even start this question because that's how unsure I am.

- jtvatsim

Alright, that's fine. Just a quick review then. With function compositions remember that one function "eats" the other one. What on earth am I talking about? An example would help...
Think about two easier functions just for practice. \[f(x) = \sqrt{x}\] and \[g(x) = 4-x\]
Then, if we are to find \[f \circ g\], we will plug in all of g(x) into the f function. We get this \[f\circ g = \sqrt{g(x)} = \sqrt{4-x}\]
Does that make sense on how the composition works?

- jtvatsim

Since sometimes I personally like to know what's coming next, for our situation we will get this
\[\sqrt{8g(x)} = \sqrt{8 \cdot \frac{1}{x-4}} = \sqrt{\frac{8}{x-4}}\]

- jtvatsim

Basically, replace the "x" in the original f function with the "g(x)" function. That's all there is to finding the composition.

- jtvatsim

Let me know when you are ready to look at the domain part. :)

- anonymous

I'm a little overwhelmed by this..

- jtvatsim

I imagine. It's quite a lot to get hit with. Maybe we should start simpler. Let's think about the functions from the actual problem.

- jtvatsim

\[f(x) = \sqrt{8x}\] and \[g(x) = \frac{1}{x-4}\]
We're ok with that so far right?

- anonymous

Yes, this is a nice start.

- jtvatsim

OK, now let's get the composition thing straight, maybe my first attempt was too complex. If I told you to find f(2) that would be easy right?

- anonymous

I apologize, but where would I even began to do that?

- jtvatsim

OK! Great question. No need to apologize. Seems like you are a little confused on function notation. That's fine, I was before too. :)
f(2) is a shorthand way of saying "Plug x = 2 into the function for f."

- jtvatsim

So, what we would see is \[f(2) = \sqrt{8\cdot 2} = \sqrt{16}.\] It's really not that bad once you know what the "secret code" is. Any questions on that?

- anonymous

Wow that made things a lot simplier. Okay we're getting somewhere.

- jtvatsim

Haha! I always had those moments in math class, "Why didn't you just say so???" :)

- anonymous

Haha :)

- jtvatsim

Glad to help! No need to be in the dark about the secret of functions. They are kind of dumb, but some mathematician way back in the 1700s or something created a shorthand code for himself, and everybody just started copying him.

- jtvatsim

So, anyways. It's easy to plug in a single number into a function. We could do
f(2)
f(1)
f(100)
g(45)
g(1)
g(3.4)
until the cows came home. This is easy. We just plug x = some number into the formula.

- jtvatsim

Now, the big word "composition" sounds like really sophisticated rocket science. But it's actually very simple and is based on plugging in numbers.

- jtvatsim

A "composition" just happens when we plug an "equation" into the function instead of a single "number".

- jtvatsim

It's weird, but it's the same idea as plugging a number into the function.

- jtvatsim

So, if I said f(2x-4), I have done a composition. I plugged in an equation instead of a number. I would do the same thing as if it were a number:
\[f(2x-4) = \sqrt{8\cdot(2x-4)} = \sqrt{16x - 32}\]
That's a lot at once, but do you kind of see how that works?

- jtvatsim

By the way, once you have these basic concepts down, you'll never have to do this depth of thinking again. The question itself is very easy, but it assumes that you know about composition. :)

- anonymous

Ok, I'm starting to understand better now

- jtvatsim

"basic" as in "the foundation of the question" , not "basic" as in "duh easy". :)

- jtvatsim

Alright, good! We are almost ready to finish off the problem itself. There's just one more "shorthand" AGAIN to mention.

- jtvatsim

Since functions (basically the same thing as equations) sometimes have names (like f or g) we could also write things like
f(g(x))
What this means is plug whatever the g(x) equation is into the f function.

- jtvatsim

So, if g(x) were something like g(x) = 3x. Then,
\[f(g(x)) = f(3x) = \sqrt{8 \cdot (3x)}\]

- jtvatsim

Obviously, I'm using a different g(x) than the problem as an example, but does that part make sense so far? That's basically what we did a minute ago.

- anonymous

Yes it is definitely starting to make sense.

- jtvatsim

OK, cool. You are doing great!

- jtvatsim

Alright, so mathematicians like some messes (just look at some equations), but don't like others (they're weird like that). They decided that they don't like the way that this
f(g(x))
looks. It has too many parentheses and looks messy.

- jtvatsim

I mean, we just used it and it makes sense... but nooooooo.... they created a new "shorthand" to deal with this:
\[f \circ g = f(g(x))\]

- jtvatsim

So, what they did is to say "Alright, when we write f circle thing g" we REALLY mean "plug the g(x) equation into the f function".... whatever, it's a code, it's fine, it's dumb... but yeah, sure, whatever you say. :P

- jtvatsim

That's what the "f circle thing g" means in the question. Just thought I would point that out. OK so far?

- anonymous

So fog is just another name for f(g(x))?

- jtvatsim

Yes. Why do they have to confuse the issue? I don't know... And by the same logic:
gof is just another name for g(f(x)) [if for some reason you wanted to plug f into g]

- anonymous

I'm glad you're helping me with this, I would of never knew otherwise.

- jtvatsim

Yep, it only took me getting a math major at university for me to start figuring this stuff out... how would we have known?

- jtvatsim

In any case, now that we know what f(number) means "just plug x = number" and we know that composition is just plugging in an equation, we can now start answering the crazy (but relatively easy) question given to us.

- anonymous

Let's get it :)

- jtvatsim

We are told to think about fog, we now know that this means f(g(x)). So, just plug in the equation for g into the one for f.
\[f\circ g = f(g(x)) = f(\frac{1}{x-4}) = \sqrt{8 \cdot \frac{1}{x-4}} = \sqrt{\frac{8}{x-4}}\]

- jtvatsim

He's an ugly looking fella isn't he? But, that can't be helped... :)

- anonymous

This looks a lot better seeing it now then when we first started.

- jtvatsim

Yep, funny how the exact equation can look like total gibberish one second, then actually look simpler once you know the meanings behind all the symbols.

- jtvatsim

So, now for the next idea: Domain.... oooooooohhh aaaaahhh. Nope. Not that impressive.
Domains are just lists of numbers that are allowed in the equation. There are only a few things that CANNOT ever HAPPEN:
1. We do not allow fractions to divide by 0.
2. We do not allow square roots to have negative insides. Like \[\sqrt{-4}\] is not allowed.
3. There are other "bad things." If you study logarithms (sort of like exponents), we don't allow them to equal 0 or be negative. Don't worry about this for now, we don't have logarithms in this question.

- jtvatsim

The question wants us to find the number that is definitely in the domain (it is an OK number and doesn't break any rules...)

- jtvatsim

Start with 4. Does plugging in x = 4 into our crazy equation cause any problems?

- anonymous

It doesn't seem like it.

- anonymous

Of course I may be wrong.

- jtvatsim

Sure, and that's totally fine, let's take a close look at the "instant reply" :)

- jtvatsim

\[\sqrt{\frac{8}{x-4}} \rightarrow \sqrt{\frac{8}{4-4}} \rightarrow \sqrt{\frac{8}{0}}\]
duhn duhn... DUUUUUHHNNN!

- jtvatsim

OH NO! It's the revenge of dividing a fraction by 0!

- jtvatsim

So, our poor 4 is not in the domain. He creates an equation that doesn't make sense to us. The answer is not 4. Any questions on that one?

- anonymous

Wow, I see how this works now.

- jtvatsim

Awesome! We will continue this little plug-in game until we find something that works.
How about x = 0 as our next candidate?

- anonymous

0's aren't allowed, if I'm not mistaken.

- jtvatsim

0's aren't allowed to be divided by. 0's in general are OK.
So, this is OK: 0 + 4. Sure I know what that means.
And this is OK: 8 * 0. I know what that means.
Even this is OK: 0/8. That's just 0.
But dividing by 0 is not OK: 8/0. DANGER DANGER!
So, let's just plug 0 in and see what happens. We never know.

- anonymous

Ohhhhh

- jtvatsim

\[\sqrt{\frac{8}{0-4}} = \sqrt{\frac{8}{-4}}\] so far so good.

- jtvatsim

\[\sqrt{-2}\]
aww... dang it...

- jtvatsim

negative insides in square roots are not allowed. x = 0 fails the test... at least he didn't divide by 0, but he's still out.

- anonymous

I think 6 is our number for this one.

- jtvatsim

Alright, let's see, I really hope so. x = 6... come on down, you are our next contestant!

- jtvatsim

\[\sqrt{\frac{8}{6-4}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\] VICTORY!

- jtvatsim

Finally, something that makes sense! You can see that x = -9 will be bad news just like x = 0. it will create a square root of a negative number. We have our answer x = 6 is in the domain and A-OKAY!

- anonymous

You make math pretty fun to learn, and I don't even know how that's possible.

- jtvatsim

LOL! Maybe when you've been tortured enough with math you just finally crack and start making sense. I'm glad you found it enjoyable! If only more teachers would see that math has real meaning, it's not just random rules (well... some rules are random). It doesn't need to be overly complicated. :)

- anonymous

Yeah, if only they knew. Lol.

- anonymous

Okay so I have another question similar to this one

- jtvatsim

Alright, I'll take a quick look, I may have to check out soon, it's almost midnight here and as you can probably tell I'm clearly starting to lose it. But's let's take a look. :)

- jtvatsim

This thread is getting cluttered and slowing down, so if you could post a new question that'd be great. :)

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