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anonymous

  • one year ago

Domain Values question. (Picture of question in thread)

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @jtvatsim

  3. jtvatsim
    • one year ago
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    K, let's see...

  4. jtvatsim
    • one year ago
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    First of all, we should write down what the composition looks like, it's a bit ugly, but have you done that step?

  5. anonymous
    • one year ago
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    No, I didn't even start this question because that's how unsure I am.

  6. jtvatsim
    • one year ago
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    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?

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

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

  9. jtvatsim
    • one year ago
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    Let me know when you are ready to look at the domain part. :)

  10. anonymous
    • one year ago
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    I'm a little overwhelmed by this..

  11. jtvatsim
    • one year ago
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    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.

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

  13. anonymous
    • one year ago
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    Yes, this is a nice start.

  14. jtvatsim
    • one year ago
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    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?

  15. anonymous
    • one year ago
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    I apologize, but where would I even began to do that?

  16. jtvatsim
    • one year ago
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    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."

  17. jtvatsim
    • one year ago
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    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?

  18. anonymous
    • one year ago
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    Wow that made things a lot simplier. Okay we're getting somewhere.

  19. jtvatsim
    • one year ago
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    Haha! I always had those moments in math class, "Why didn't you just say so???" :)

  20. anonymous
    • one year ago
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    Haha :)

  21. jtvatsim
    • one year ago
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    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.

  22. jtvatsim
    • one year ago
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    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.

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

  24. jtvatsim
    • one year ago
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    A "composition" just happens when we plug an "equation" into the function instead of a single "number".

  25. jtvatsim
    • one year ago
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    It's weird, but it's the same idea as plugging a number into the function.

  26. jtvatsim
    • one year ago
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    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?

  27. jtvatsim
    • one year ago
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    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. :)

  28. anonymous
    • one year ago
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    Ok, I'm starting to understand better now

  29. jtvatsim
    • one year ago
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    "basic" as in "the foundation of the question" , not "basic" as in "duh easy". :)

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

  31. jtvatsim
    • one year ago
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    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.

  32. jtvatsim
    • one year ago
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    So, if g(x) were something like g(x) = 3x. Then, \[f(g(x)) = f(3x) = \sqrt{8 \cdot (3x)}\]

  33. jtvatsim
    • one year ago
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    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.

  34. anonymous
    • one year ago
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    Yes it is definitely starting to make sense.

  35. jtvatsim
    • one year ago
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    OK, cool. You are doing great!

  36. jtvatsim
    • one year ago
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    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.

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

  38. jtvatsim
    • one year ago
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    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

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

  40. anonymous
    • one year ago
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    So fog is just another name for f(g(x))?

  41. jtvatsim
    • one year ago
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    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]

  42. anonymous
    • one year ago
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    I'm glad you're helping me with this, I would of never knew otherwise.

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

  44. jtvatsim
    • one year ago
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    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.

  45. anonymous
    • one year ago
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    Let's get it :)

  46. jtvatsim
    • one year ago
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    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. \[f\circ g = f(g(x)) = f(\frac{1}{x-4}) = \sqrt{8 \cdot \frac{1}{x-4}} = \sqrt{\frac{8}{x-4}}\]

  47. jtvatsim
    • one year ago
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    He's an ugly looking fella isn't he? But, that can't be helped... :)

  48. anonymous
    • one year ago
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    This looks a lot better seeing it now then when we first started.

  49. jtvatsim
    • one year ago
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    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.

  50. jtvatsim
    • one year ago
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    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.

  51. jtvatsim
    • one year ago
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    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...)

  52. jtvatsim
    • one year ago
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    Start with 4. Does plugging in x = 4 into our crazy equation cause any problems?

  53. anonymous
    • one year ago
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    It doesn't seem like it.

  54. anonymous
    • one year ago
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    Of course I may be wrong.

  55. jtvatsim
    • one year ago
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    Sure, and that's totally fine, let's take a close look at the "instant reply" :)

  56. jtvatsim
    • one year ago
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    \[\sqrt{\frac{8}{x-4}} \rightarrow \sqrt{\frac{8}{4-4}} \rightarrow \sqrt{\frac{8}{0}}\] duhn duhn... DUUUUUHHNNN!

  57. jtvatsim
    • one year ago
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    OH NO! It's the revenge of dividing a fraction by 0!

  58. jtvatsim
    • one year ago
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    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?

  59. anonymous
    • one year ago
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    Wow, I see how this works now.

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

  61. anonymous
    • one year ago
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    0's aren't allowed, if I'm not mistaken.

  62. jtvatsim
    • one year ago
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    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.

  63. anonymous
    • one year ago
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    Ohhhhh

  64. jtvatsim
    • one year ago
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    \[\sqrt{\frac{8}{0-4}} = \sqrt{\frac{8}{-4}}\] so far so good.

  65. jtvatsim
    • one year ago
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    \[\sqrt{-2}\] aww... dang it...

  66. jtvatsim
    • one year ago
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    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.

  67. anonymous
    • one year ago
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    I think 6 is our number for this one.

  68. jtvatsim
    • one year ago
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    Alright, let's see, I really hope so. x = 6... come on down, you are our next contestant!

  69. jtvatsim
    • one year ago
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    \[\sqrt{\frac{8}{6-4}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\] VICTORY!

  70. jtvatsim
    • one year ago
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    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!

  71. anonymous
    • one year ago
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    You make math pretty fun to learn, and I don't even know how that's possible.

  72. jtvatsim
    • one year ago
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    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. :)

  73. anonymous
    • one year ago
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    Yeah, if only they knew. Lol.

  74. anonymous
    • one year ago
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    Okay so I have another question similar to this one

  75. jtvatsim
    • one year ago
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    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. :)

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

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