Question

Domain Values question. (Picture of question in thread)

Answer 1

@jtvatsim

Answer 2

K, let's see...

Answer 3

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

Answer 4

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

Answer 5

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?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

I'm a little overwhelmed by this..

Answer 10

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.

Answer 11

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

Answer 12

Yes, this is a nice start.

Answer 13

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?

Answer 14

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

Answer 15

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."

Answer 16

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?

Answer 17

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

Answer 18

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

Answer 19

Haha :)

Answer 20

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.

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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?

Answer 26

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. :)

Answer 27

Ok, I'm starting to understand better now

Answer 28

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

Answer 29

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

Answer 30

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.

Answer 31

So, if g(x) were something like g(x) = 3x. Then,

\[f(g(x)) = f(3x) = \sqrt{8 \cdot (3x)}\]

Answer 32

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.

Answer 33

Yes it is definitely starting to make sense.

Answer 34

OK, cool. You are doing great!

Answer 35

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.

Answer 36

I mean, we just used it and it makes sense... but nooooooo.... they created a new "shorthand" to deal with this:

\[f \circ g = f(g(x))\]

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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]

Answer 41

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

Answer 42

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

Answer 43

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.

Answer 44

Let's get it :)

Answer 45

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.

\[f\circ g = f(g(x)) = f(\frac{1}{x-4}) = \sqrt{8 \cdot \frac{1}{x-4}} = \sqrt{\frac{8}{x-4}}\]

Answer 46

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

Answer 47

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

Answer 48

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.

Answer 49

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.

Answer 50

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...)

Answer 51

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

Answer 52

It doesn't seem like it.

Answer 53

Of course I may be wrong.

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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?

Answer 58

Wow, I see how this works now.

Answer 59

Awesome! We will continue this little plug-in game until we find something that works.

How about x = 0 as our next candidate?

Answer 60

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

Answer 61

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.

Answer 62

Ohhhhh

Answer 63

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

Answer 64

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

Answer 65

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.

Answer 66

I think 6 is our number for this one.

Answer 67

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

Answer 68

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

Answer 69

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!

Answer 70

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

Answer 71

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. :)

Answer 72

Yeah, if only they knew. Lol.

Answer 73

Okay so I have another question similar to this one

Answer 74

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. :)

Answer 75

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