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Issy14

  • one year ago

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.

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  1. geerky42
    • one year ago
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    well, we need to know what g(x) is. were you given function of g?

  2. Issy14
    • one year ago
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    @geerky42 one second going to post the image

  3. Issy14
    • one year ago
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    |dw:1404857483770:dw|

  4. Issy14
    • one year ago
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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.

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

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

  7. Issy14
    • one year ago
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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

  8. Issy14
    • one year ago
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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)

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

  10. Issy14
    • one year ago
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    help @jdoe0001

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

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

  13. Issy14
    • one year ago
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    with the problem given

  14. Issy14
    • one year ago
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    it comes with the graph and the instructions.

  15. Issy14
    • one year ago
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    How did you find f(x). ? did you just transform it from the original function |x|

  16. Issy14
    • one year ago
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    @geerky42

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

  18. Issy14
    • one year ago
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    alright, is that the usual approach?

  19. Issy14
    • one year ago
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    am I correct on g(x)?

  20. geerky42
    • one year ago
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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?

  21. Issy14
    • one year ago
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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)

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

  23. geerky42
    • one year ago
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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 \(f'(x) = h'(~j(x)~)~j'(x)\)

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

  25. geerky42
    • one year ago
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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?

  26. geerky42
    • one year ago
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    what does "estimate the derivative" mean?

  27. Issy14
    • one year ago
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    basically, when they ask us to do that they want an approximate value not the absolute value

  28. geerky42
    • one year ago
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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.

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

  30. satellite73
    • one year ago
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    i think you are thinking too hard for this one

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

  32. satellite73
    • one year ago
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    similarly \(g(2)=2\) so \(g(g(2))=g(2)=2\)

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

  34. satellite73
    • one year ago
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    ok that was a mistake!!

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

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

  37. febylailani
    • one year ago
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    |dw:1404869501887:dw|

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

  39. Issy14
    • one year ago
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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

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