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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.

  • one month ago
  • one month 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?

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

    • one month ago
  3. Issy14 Group Title
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    |dw:1404857483770:dw|

    • one month 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.

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

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

    • one month 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

    • one month 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)

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

    • one month ago
  10. Issy14 Group Title
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    help @jdoe0001

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

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

    • one month ago
  13. Issy14 Group Title
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    with the problem given

    • one month ago
  14. Issy14 Group Title
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    it comes with the graph and the instructions.

    • one month 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|

    • one month ago
  16. Issy14 Group Title
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    @geerky42

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

    • one month ago
  18. Issy14 Group Title
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    alright, is that the usual approach?

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

    • one month 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?

    • one month 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)

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

    • one month 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 \(f'(x) = h'(~j(x)~)~j'(x)\)

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

    • one month 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?

    • one month ago
  26. geerky42 Group Title
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    what does "estimate the derivative" mean?

    • one month 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

    • one month 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.

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

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

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

    • one month ago
  32. satellite73 Group Title
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    similarly \(g(2)=2\) so \(g(g(2))=g(2)=2\)

    • one month 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

    • one month ago
  34. satellite73 Group Title
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    ok that was a mistake!!

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

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

    • one month ago
  37. febylailani Group Title
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    |dw:1404869501887:dw|

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

    • one month 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

    • one month ago
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