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anonymous

  • 4 years ago

If f and g are the functions whose graphs are shown below, let u(x)=f(g(x)) and v(x)=g(f(x)) . find u'(3) and v'(3)

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  1. anonymous
    • 4 years ago
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  2. TuringTest
    • 4 years ago
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    start with finding expressions for u'(x) and v'(x)

  3. TuringTest
    • 4 years ago
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    u'(x)=f'(g(x))g'(x) v'(x)=g'(f(x))f'(x)

  4. anonymous
    • 4 years ago
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    yeah i dont understand this. am i just looking at 3 on the graph?

  5. TuringTest
    • 4 years ago
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    say you want u'(3), you first need g(3) because g(x) is part of the formula you need what is g(3) according to the graph?

  6. anonymous
    • 4 years ago
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    looks like 4

  7. TuringTest
    • 4 years ago
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    looks like g(3)=10 to me each square is 2 vertically, and 1 horizontally as far as I can tell

  8. TuringTest
    • 4 years ago
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    g(3)=8 I mean, sorry

  9. TuringTest
    • 4 years ago
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    I'm not really sure about the increments on the graph though, it's a bit vague...

  10. anonymous
    • 4 years ago
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    8, how so?

  11. TuringTest
    • 4 years ago
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    It looks to me like each square vertically is 2 (by the marking), but I guess it must be 4 since 8 is not on the graph horizontally so ok, g(3)=4 what is the value of f'(4) ? [remember we need f'(g(3))g'(3)]

  12. TuringTest
    • 4 years ago
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    also remember that f'(4) represents the slope of f(x) at x=4

  13. anonymous
    • 4 years ago
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    f(4) is 2

  14. TuringTest
    • 4 years ago
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    we want f'(4) that is the slope of f(x) at x=4 what is the slope of the line in the graph f(x) at x=4 ?

  15. anonymous
    • 4 years ago
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    when x is 4, f is 2 right?

  16. TuringTest
    • 4 years ago
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    yes, but that is the value of f(4) not f'(4) f'(4) is the SLOPE, not the value of the function do you remember how to find the slope of a straight line from algebra?

  17. anonymous
    • 4 years ago
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    ohh

  18. anonymous
    • 4 years ago
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    y2-y1/x2-x1?

  19. TuringTest
    • 4 years ago
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    actually sorry, they are the same in this case, but that is pure luck! lol notice that the slope (rise over run) is also 2 !

  20. TuringTest
    • 4 years ago
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    that is what f' means, the slope of f so similarly what is the last piece of the formula we need ? u'(3)=f'(g(3))g'(3) and we still need g'(3) what is it?

  21. anonymous
    • 4 years ago
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    rise over run right? 4/3? no?

  22. TuringTest
    • 4 years ago
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    at x=3 it looks to me like g has a slope of -1

  23. anonymous
    • 4 years ago
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    how did you figure that out?

  24. TuringTest
    • 4 years ago
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    look at the tail last portion of g(x), the far right portion of the top graph. it dips downward it seems to go down one unit for every unit it goes to the right rise/run=-1/1=-1

  25. TuringTest
    • 4 years ago
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    here's a sketch of g|dw:1329536578080:dw|

  26. anonymous
    • 4 years ago
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    ok i see now i see

  27. TuringTest
    • 4 years ago
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    here's the part from x=2 to x=4

  28. TuringTest
    • 4 years ago
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    |dw:1329536693848:dw|in your graph you can see it goes one over and one down on that region

  29. TuringTest
    • 4 years ago
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    (I'm not gonna draw the little squares in...)

  30. anonymous
    • 4 years ago
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    i see that thanks

  31. TuringTest
    • 4 years ago
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    so you have all the pieces now u'(3)=f'(g(3))g'(3) you know f'(g(3)) and g'(3), so multiply them to find u'(3)

  32. TuringTest
    • 4 years ago
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    what is g(3) ?

  33. anonymous
    • 4 years ago
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    4 slope is -1

  34. TuringTest
    • 4 years ago
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    right, g(3)=4 and what is f'(g(3)) ?

  35. anonymous
    • 4 years ago
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    f is also 4 right? so that makes 4 x 4?

  36. TuringTest
    • 4 years ago
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    no, we want f'(g(3))=f'(4) because g(3)=3, right ? what is f'(4) ?

  37. anonymous
    • 4 years ago
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    the slope of at 4 is 3

  38. TuringTest
    • 4 years ago
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    you are thinking about the idea correctly, but it looks like 2 to me

  39. anonymous
    • 4 years ago
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    ok the gradients on this graph is crap

  40. TuringTest
    • 4 years ago
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    yeah, I agree with that, but the problem only seems to make sense if we count each square as 1...|dw:1329537340258:dw|so we want the slope of the upward portion of f

  41. anonymous
    • 4 years ago
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    4/2

  42. anonymous
    • 4 years ago
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    2

  43. TuringTest
    • 4 years ago
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    right, looks that way so what is u'(3)=f'(g(3))g'(3) ?

  44. anonymous
    • 4 years ago
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    (2)(4) (4)

  45. TuringTest
    • 4 years ago
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    there should be no 4 in the answer what is f'(g(3)) ?

  46. TuringTest
    • 4 years ago
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    (we just figured it out above)

  47. anonymous
    • 4 years ago
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    2

  48. TuringTest
    • 4 years ago
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    yes, and what is g'(3) ?

  49. anonymous
    • 4 years ago
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    -1

  50. TuringTest
    • 4 years ago
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    so f'(g(3))g'(3) is what?

  51. anonymous
    • 4 years ago
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    (2) (-1)

  52. TuringTest
    • 4 years ago
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    u'(3)=f'(g(3))g'(3)=-2 yep :D

  53. anonymous
    • 4 years ago
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    thanks for the help

  54. TuringTest
    • 4 years ago
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    so now you get to try for v'(3) ! good luck :)

  55. anonymous
    • 4 years ago
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    I got zero

  56. TuringTest
    • 4 years ago
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    yep!

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