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znimon

  • 2 years ago

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

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

  2. LolWolf
    • 2 years ago
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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.

  3. znimon
    • 2 years ago
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    This is pre-calc

  4. LolWolf
    • 2 years ago
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    All right, then that should be the case.

  5. znimon
    • 2 years ago
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    so you are telling me my answer is right?

  6. znimon
    • 2 years ago
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    /correct

  7. LolWolf
    • 2 years ago
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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.

  8. znimon
    • 2 years ago
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    how did you get -7h out from under the root?

  9. LolWolf
    • 2 years ago
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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.

  10. LolWolf
    • 2 years ago
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    This is only useful for evaluating the limit.

  11. znimon
    • 2 years ago
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    I have no idea what you mean by that

  12. znimon
    • 2 years ago
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    why would I multiply the numerator and denominator by that?

  13. znimon
    • 2 years ago
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    @LolWolf

  14. znimon
    • 2 years ago
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    My 7 sub 2 is \[\sqrt{21-7h}\]

  15. znimon
    • 2 years ago
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    Y sub 2*

  16. LolWolf
    • 2 years ago
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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.

  17. znimon
    • 2 years ago
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    yours simplifies to -7

  18. znimon
    • 2 years ago
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    f(-3+h) = \[\sqrt{-7(3+h)}\]

  19. LolWolf
    • 2 years ago
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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).

  20. znimon
    • 2 years ago
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    The original problem looks like the square root goes over the "t"

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

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

  23. znimon
    • 2 years ago
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    @LolWolf

  24. LolWolf
    • 2 years ago
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    Where is there not a square root sign?

  25. znimon
    • 2 years ago
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    Over the "7h"

  26. znimon
    • 2 years ago
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    the 7h that is in the numerator of your third comment

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

  28. znimon
    • 2 years ago
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    oh that is weird it doesn't look like that in my browser

  29. znimon
    • 2 years ago
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    so my first comment is correct then

  30. znimon
    • 2 years ago
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    Square root of (-7h) divided by h

  31. LolWolf
    • 2 years ago
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    No, it is not, as they are not equivalent statements.

  32. znimon
    • 2 years ago
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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?

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

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

  35. LolWolf
    • 2 years ago
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    Nope. I don't think there is any need to, unless you're taking limits.

  36. znimon
    • 2 years ago
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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?

  37. znimon
    • 2 years ago
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    Never mind I just proved it.

  38. znimon
    • 2 years ago
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    Thanks again I'll have to look up a khan academy video on that

  39. LolWolf
    • 2 years ago
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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.

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