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Issy14 Group Title

Let w(x) = g(g(x)). Find: (a) w(1) (b) w(2) (c) w(3) somebody please help, I have no idea what to do.

  • 17 days ago
  • 17 days ago

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  1. geerky42 Group Title
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    well, we need to know what g(x) is. were you given function of g?

    • 17 days ago
  2. Issy14 Group Title
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    @geerky42 one second going to post the image

    • 17 days ago
  3. Issy14 Group Title
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    |dw:1404857483770:dw|

    • 17 days ago
  4. Issy14 Group Title
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    These are the directions: In Problems 57–60, use Figure 3.17 and the chain rule to estimate the derivative, or state why the chain rule does not apply. The graph of f(x) has a sharp corner at x = 2.

    • 17 days ago
  5. Issy14 Group Title
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    If you have time , I could really use assistance on this question. @TuringTest

    • 17 days ago
  6. Issy14 Group Title
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    @tkhunny If you have time, I could really use assistance on this question

    • 17 days ago
  7. Issy14 Group Title
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    I think i figured out g(x) = -1x+4 f(x) is not continuous it has a breat at (0,2) therefore if i try to get the slope they are going to have the same magnitude but different signs f(x)=4x or f(x)= -4x

    • 17 days ago
  8. Issy14 Group Title
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    I know the chain rule by heart, I don't know how to apply here. f'(x)= f'(g(x)) x g'(x)

    • 17 days ago
  9. Issy14 Group Title
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    Help in need of a mentor please. @Taylor<3sRin @myko @ganeshie8

    • 17 days ago
  10. Issy14 Group Title
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    help @jdoe0001

    • 17 days ago
  11. geerky42 Group Title
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    \(f(x) = -2|x-2|+4\)

    • 17 days ago
  12. geerky42 Group Title
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    hmm, wait. exactly what do you need help with?

    • 17 days ago
  13. Issy14 Group Title
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    with the problem given

    • 17 days ago
  14. Issy14 Group Title
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    it comes with the graph and the instructions.

    • 17 days ago
  15. Issy14 Group Title
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    How did you find f(x). ? did you just transform it from the original function |x|

    • 17 days ago
  16. Issy14 Group Title
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    @geerky42

    • 17 days ago
  17. geerky42 Group Title
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    yeah, that's how I figure it out

    • 17 days ago
  18. Issy14 Group Title
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    alright, is that the usual approach?

    • 17 days ago
  19. Issy14 Group Title
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    am I correct on g(x)?

    • 17 days ago
  20. geerky42 Group Title
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    well, i don't know what's your given problems are. are you asked to take derivative of f(x) using chain rule, or state why chain rule does not apply?

    • 17 days ago
  21. Issy14 Group Title
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    both In Problems 57–60, use Figure 3.17 and the chain rule to estimate the derivative, or state why the chain rule does not apply. The graph of f(x) has a sharp corner at x = 2. 60. Let w(x) = g(g(x)). Find: (a) w(1) (b) w(2) (c) w(3)

    • 17 days ago
  22. Issy14 Group Title
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    that's the problem and then they give you the graph

    • 17 days ago
  23. geerky42 Group Title
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    ok i see. well, we managed to figure f(x) out by using transformation from parent function, so we have f(x) = -2|x-2|+4 we can take derivative of f(x) because it is continuous. just not differentiable at break point. so we can let h(x) = -2x+4 and j(x) = |x-2| so we have \(f'(x) = h'(~j(x)~)~j'(x)\)

    • 17 days ago
  24. geerky42 Group Title
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    PS: \[\large\dfrac{d}{dx}|x| = \dfrac{x}{|x|}\]

    • 17 days ago
  25. geerky42 Group Title
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    on second thought, you was asked to estimate the derivative, but what we are doing is finding the actual derivative. hmm, not sure what to do here. do you have any idea?

    • 17 days ago
  26. geerky42 Group Title
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    what does "estimate the derivative" mean?

    • 17 days ago
  27. Issy14 Group Title
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    basically, when they ask us to do that they want an approximate value not the absolute value

    • 17 days ago
  28. geerky42 Group Title
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    huh, i am sorry, this problem confused me, because i don't see how chain rule can be apply to estimate value from given graph? so i am not sure... sorry.

    • 17 days ago
  29. Issy14 Group Title
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    it's ok, I'm confused as well, thank you for your help.

    • 17 days ago
  30. satellite73 Group Title
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    i think you are thinking too hard for this one

    • 16 days ago
  31. satellite73 Group Title
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    from the picture we see \(g(1)=3\) and \(g(3)=1\) so \(g(g(1))=g(3)=1\)

    • 16 days ago
  32. satellite73 Group Title
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    similarly \(g(2)=2\) so \(g(g(2))=g(2)=2\)

    • 16 days ago
  33. satellite73 Group Title
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    as for the derivative, \[\left(g(g(x)\right)'=g(g(x))g'(x)\] by the chain rule

    • 16 days ago
  34. satellite73 Group Title
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    ok that was a mistake!!

    • 16 days ago
  35. satellite73 Group Title
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    \[\left(g(g(x)\right)'=g'(g(x))g'(x)\]

    • 16 days ago
  36. satellite73 Group Title
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    then \[w'(1)=g'(g(1))\times g'(1)=-1\times -1=1\]

    • 16 days ago
  37. febylailani Group Title
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    |dw:1404869501887:dw|

    • 16 days ago
  38. satellite73 Group Title
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    which should not surprise you since \(g(g(x))=x\)

    • 16 days ago
  39. Issy14 Group Title
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    then there was no need to find out the function of each one? I did that before but I thought that it was too simple. I will continue reading your answer, thank you. @satellite73 and @febylailani

    • 16 days ago
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