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anonymous

  • 5 years ago

For position function s(t):=20-5t^2, if h is a small number, find the average rate of change of s(t) on the interval [2,2+h]. (Simplify fully)

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  1. amistre64
    • 5 years ago
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    ug ..... i spose

  2. anonymous
    • 5 years ago
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    ?

  3. amistre64
    • 5 years ago
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    s(2+h) -s(2) ----------- h

  4. amistre64
    • 5 years ago
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    its alot of work just to prove that derivatives work ....

  5. anonymous
    • 5 years ago
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    lol true.

  6. amistre64
    • 5 years ago
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    20-5(2+h)^2 -(20-5(2)^2) ------------------------ h 20-5(4 +4h +h^2) -20 +20 ------------------------ h 20-20-20h-5h^2 ---------------- h -20h-5h^2 ---------- = -20 -5h h

  7. anonymous
    • 5 years ago
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    I think I have to use the formula (f(a)-f(x))/(a-x)

  8. anonymous
    • 5 years ago
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    Can you explain to me how to use that formula?

  9. amistre64
    • 5 years ago
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    now for the derivative we get: -10t ; at t=2 we get -20 when h=0, we get -20

  10. amistre64
    • 5 years ago
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    its the same formula

  11. anonymous
    • 5 years ago
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    oh no derivatives yet

  12. anonymous
    • 5 years ago
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    can you explain to me how to use it? like what's a and what's x?

  13. anonymous
    • 5 years ago
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    on the interval

  14. amistre64
    • 5 years ago
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    yes, you are doing the proof for derivatives so that in the next few weeks your teacher will say, and this is how we do it the easy way...

  15. anonymous
    • 5 years ago
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    haha ok.

  16. amistre64
    • 5 years ago
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    f(a) and f(x) simply mean to use the formula with an a and an x instead of a t like this: f(a) - f(x) (20-5a^2) - (20-5x^2) ------------------- a-x

  17. amistre64
    • 5 years ago
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    20-5a^2 -20+5x^2 ----------------- a - x -5a^2+5x^2 ------------- a - x -5(a^2-x^2) ------------- a - x -5(a+x)(a-x) ------------- a - x -5(a+x)

  18. anonymous
    • 5 years ago
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    ok, what do you plug into f(a)?

  19. anonymous
    • 5 years ago
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    in [2,2+h]

  20. amistre64
    • 5 years ago
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    .............. a = 2 and x = h

  21. anonymous
    • 5 years ago
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    ?

  22. anonymous
    • 5 years ago
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    is a=2 from the 2+h part or from just the 2 part.

  23. amistre64
    • 5 years ago
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    it doesnt matter..... a =2 OR a=h, OR x=2 OR x=h.... -5(a+x) -5(2+h) is the same things as -5(h+2)

  24. anonymous
    • 5 years ago
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    I'm confused.

  25. anonymous
    • 5 years ago
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    Like, I know the forumla, I just don't understand how to use it.

  26. amistre64
    • 5 years ago
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    then dont use it lol

  27. amistre64
    • 5 years ago
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    does it matter if f(a) = 2 and f(x) = (2+h)? no

  28. amistre64
    • 5 years ago
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    becasuse when h=0 it just disappears to -20

  29. amistre64
    • 5 years ago
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    \[\frac{y-y_0}{x-x_0}\] do you understand this formula?

  30. anonymous
    • 5 years ago
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    yes.

  31. amistre64
    • 5 years ago
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    that is all it is; but they rename that parts ..... but it doesnt matter what they name them, they still do the same things

  32. anonymous
    • 5 years ago
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    so the bottom part is a-x... don't you put 2-h?

  33. anonymous
    • 5 years ago
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    why did you only put h?

  34. amistre64
    • 5 years ago
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    \[\frac{f(a)-f(x)}{a-x} \iff \frac{y-y_0}{x-x_0} \iff \frac{f(x+h)-f(x)}{h}\]

  35. anonymous
    • 5 years ago
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    oh ok.

  36. amistre64
    • 5 years ago
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    because \((x+h)-(x)=h\)

  37. anonymous
    • 5 years ago
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    why?

  38. amistre64
    • 5 years ago
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    why what? why does x-x+h = 0+h = h?

  39. anonymous
    • 5 years ago
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    oh ok. just kidding.

  40. amistre64
    • 5 years ago
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    \[(x+h)-(x)\] \[x+h-x\] \[x-x+h\] \[(x-x)+h\] \[0+h\] \[h\]

  41. anonymous
    • 5 years ago
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    so the formula would be \[(f(2+h)-f(2))\div(h)\]

  42. amistre64
    • 5 years ago
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    yes

  43. anonymous
    • 5 years ago
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    ok.

  44. anonymous
    • 5 years ago
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    so its ((20-5(2+h)^2)-(20-5(2)^2))/h

  45. amistre64
    • 5 years ago
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    yes

  46. anonymous
    • 5 years ago
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    so (20-5(2+h)^2)/h

  47. anonymous
    • 5 years ago
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    and finally -20-h?

  48. amistre64
    • 5 years ago
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    yes

  49. anonymous
    • 5 years ago
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    :) Thank you

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