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znimon Group Title

A function is given below. Determine the average rate of change of the function between x = -3 and x = -3 + h. f(t) = √-7t

  • one year ago
  • one year ago

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  1. znimon Group Title
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    I have \[\sqrt{-7h}/h\] but it says that is wrong and I can't figure out why

    • one year ago
  2. LolWolf Group Title
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    Well, the 'average' rate of change for some interval \([a,b]\) (non-calculus, please tell me if you need otherwise) would be: \[ \Delta f_{avg}=\frac{f(b)-f(a)}{b-a} \]Try using that.

    • one year ago
  3. znimon Group Title
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    This is pre-calc

    • one year ago
  4. LolWolf Group Title
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    All right, then that should be the case.

    • one year ago
  5. znimon Group Title
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    so you are telling me my answer is right?

    • one year ago
  6. znimon Group Title
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    /correct

    • one year ago
  7. LolWolf Group Title
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    Nope, sorry. Using the above, we find, for \(f(t)=\sqrt{-7t}\) \[ \frac{f(-3+h)-f(-3)}{-3+h+3}=\frac{\sqrt{21-7h}-\sqrt{21}}{h} \]If you need further simplification of the above, please tell me.

    • one year ago
  8. znimon Group Title
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    how did you get -7h out from under the root?

    • one year ago
  9. LolWolf Group Title
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    How does one? You can't, you'd have to multiply both the numerator and denominator by \(\sqrt{21-7h}+\sqrt{21}\), but then it would end up on the denominator.

    • one year ago
  10. LolWolf Group Title
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    This is only useful for evaluating the limit.

    • one year ago
  11. znimon Group Title
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    I have no idea what you mean by that

    • one year ago
  12. znimon Group Title
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    why would I multiply the numerator and denominator by that?

    • one year ago
  13. znimon Group Title
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    @LolWolf

    • one year ago
  14. znimon Group Title
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    My 7 sub 2 is \[\sqrt{21-7h}\]

    • one year ago
  15. znimon Group Title
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    Y sub 2*

    • one year ago
  16. LolWolf Group Title
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    If you wish to remove the \(h\) from the radical, you'd have to do that, but, of course, then the top expression ends up in the denominator. So, the point is that one cannot remove the \(h\) from such.

    • one year ago
  17. znimon Group Title
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    yours simplifies to -7

    • one year ago
  18. znimon Group Title
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    f(-3+h) = \[\sqrt{-7(3+h)}\]

    • one year ago
  19. LolWolf Group Title
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    My equation does not simplify. And, yes, that last statement is correct. Keep in mind: \[ \sqrt{a+b}-\sqrt{a}=\sqrt{b}\\ \]Is *not* necessarily true (In fact, it is mainly true if b=0 or a=0).

    • one year ago
  20. znimon Group Title
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    The original problem looks like the square root goes over the "t"

    • one year ago
  21. LolWolf Group Title
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    Yes, and that's how I computed it. What do you feel is wrong with my expression?

    • one year ago
  22. znimon Group Title
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    I don't understand how there is no square root sign over the 7h in your third comment

    • one year ago
  23. znimon Group Title
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    @LolWolf

    • one year ago
  24. LolWolf Group Title
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    Where is there not a square root sign?

    • one year ago
  25. znimon Group Title
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    Over the "7h"

    • one year ago
  26. znimon Group Title
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    the 7h that is in the numerator of your third comment

    • one year ago
  27. LolWolf Group Title
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    http://imgur.com/1fwvB This is what I have in my browser and what has been typed.

    • one year ago
  28. znimon Group Title
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    oh that is weird it doesn't look like that in my browser

    • one year ago
  29. znimon Group Title
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    so my first comment is correct then

    • one year ago
  30. znimon Group Title
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    Square root of (-7h) divided by h

    • one year ago
  31. LolWolf Group Title
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    No, it is not, as they are not equivalent statements.

    • one year ago
  32. znimon Group Title
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    yeah it is because square root of (x+y) is equal to square root of x plus square root of y right?

    • one year ago
  33. LolWolf Group Title
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    No, it does not. \[ \sqrt{a+b}\ne\sqrt{a}+\sqrt{b} \]Unless a or b is zero.

    • one year ago
  34. znimon Group Title
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    oh jesus I feel like an idiot. So then your third comment does not simplify any further in pre calc?

    • one year ago
  35. LolWolf Group Title
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    Nope. I don't think there is any need to, unless you're taking limits.

    • one year ago
  36. znimon Group Title
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    so last thing square root of x*y is equal to square root of x times the square root of y?

    • one year ago
  37. znimon Group Title
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    Never mind I just proved it.

    • one year ago
  38. znimon Group Title
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    Thanks again I'll have to look up a khan academy video on that

    • one year ago
  39. LolWolf Group Title
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    Yes, that statement is true. Since: \[ a^2=n\\ b^2=m \]So we say: \[ nm=a^2b^2=(ab)^2 \]And all right, sure thing.

    • one year ago
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