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anonymous

  • one year ago

Let f(x)=√4x and g(x) = x + 6. What's the smallest number that is in the domain of fog?

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  1. jtvatsim
    • one year ago
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    Well, at least we know how to find fog now. Check if your answer matches mine for the fog part.

  2. jtvatsim
    • one year ago
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    \[f\circ g = f(g(x)) = \sqrt{4\cdot (x+6)}=\sqrt{4x + 24}\]

  3. arindameducationusc
    • one year ago
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    sqrt(4)*(x+6)

  4. arindameducationusc
    • one year ago
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    I think its not whole underoot

  5. anonymous
    • one year ago
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    Ok we're getting somewhere.

  6. arindameducationusc
    • one year ago
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    Ya @Generalman1230 , what is it?

  7. anonymous
    • one year ago
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    Hold on, give me a sec guys.

  8. anonymous
    • one year ago
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    Sorry, I just can't wrap it around my head. @jtvatsim @arindameducationusc

  9. anonymous
    • one year ago
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    I thought I had something

  10. jtvatsim
    • one year ago
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    No worries, I think we are majoring on minors right now. The question that arindameducationusc doesn't have to do with the process I taught you. It's just a matter of "before we get carried away" let's check that we actually wrote the equation for f correctly.

  11. arindameducationusc
    • one year ago
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    @Generalman1230 , @jtvatsim will take care.... don't worry..... :)

  12. jtvatsim
    • one year ago
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    OK, let's try this from the top again. I think you are fine. :)

  13. jtvatsim
    • one year ago
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    We have \[f(x) = \sqrt{4x}\] where the 4x is completely inside the root, and \[g(x) = x+6\].

  14. jtvatsim
    • one year ago
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    We then know based on our previous conversation that \[f \circ g = f(g(x)) = f(x+6) = \sqrt{4 \cdot (x+6)} = \sqrt{4x + 24}\]

  15. jtvatsim
    • one year ago
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    That's where we left off before. Are you OK up to that step?

  16. anonymous
    • one year ago
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    Yes, I am.

  17. jtvatsim
    • one year ago
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    Excellent. Now, the question wants us to find the smallest number in the domain. How small can I get before there is a problem? Well, let's experiment! Time for science!

  18. jtvatsim
    • one year ago
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    Or even better, we can stop and think about what can and can't happen.

  19. jtvatsim
    • one year ago
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    The only issue in this situation (there are no fractions, so dividing by 0 isn't a problem) is if we get a negative number inside the square root.

  20. jtvatsim
    • one year ago
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    The smallest thing that can EVER be inside a square root is 0 then. It's the first "not-negative" number. Does that logic make sense so far?

  21. jtvatsim
    • one year ago
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    Phew... got caught in an internet black hole there, don't know what happened.

  22. anonymous
    • one year ago
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    The website crashed, I guess. I couldn't get on it for a second.

  23. jtvatsim
    • one year ago
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    Yep, it's still a little shaky even now...

  24. jtvatsim
    • one year ago
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    So, are you comfortable with the idea that the smallest thing you can have inside a square root is 0? There is one more step to do after coming to grips with this fact.

  25. anonymous
    • one year ago
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    I tried to type in yes, but the site crashed again.

  26. jtvatsim
    • one year ago
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    lol, almost there site, just hang in there.

  27. jtvatsim
    • one year ago
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    K, so I'll make this quick before it crashes again.

  28. jtvatsim
    • one year ago
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    The smallest thing that can be inside is 0. BUT... we have a HUGE equation inside right now. That's not a problem, the smallest this HUGE equation inside can be is 0, so... this means: 4x + 24 = 0 is the smallest we can go, 4x = -24 is the smallest we can go, x = -6 is the smallest we can go. x = -6 is the smallest number in the domain.

  29. anonymous
    • one year ago
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    @jtvatsim I can't thank you enough for all the help you gave throughout the past hour. I know you have to get off but could you help me with one more problem?

  30. anonymous
    • one year ago
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    If you have to go now, then that's totally understandable.

  31. jtvatsim
    • one year ago
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    Gotcha, you know I do have to run. But message me the question, and I can message you back sometime tomorrow. If it's due before then, there are plenty of other great tutors online. Good luck! You have come far! :)

  32. anonymous
    • one year ago
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    @jtvatsim Will do, thanks again.

  33. jtvatsim
    • one year ago
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    You are very welcome! Summit Mt. Math and conquer it! >:) Nights!

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