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Issy14

  • 2 years 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
    • 2 years ago
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    well, we need to know what g(x) is. were you given function of g?

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

  3. Issy14
    • 2 years ago
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    |dw:1404857483770:dw|

  4. Issy14
    • 2 years 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
    • 2 years ago
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    If you have time , I could really use assistance on this question. @TuringTest

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

  7. Issy14
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Help in need of a mentor please. @Taylor<3sRin @myko @ganeshie8

  10. Issy14
    • 2 years ago
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    help @jdoe0001

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

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

  13. Issy14
    • 2 years ago
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    with the problem given

  14. Issy14
    • 2 years ago
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    it comes with the graph and the instructions.

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

  16. Issy14
    • 2 years ago
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    @geerky42

  17. geerky42
    • 2 years ago
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    yeah, that's how I figure it out

  18. Issy14
    • 2 years ago
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    alright, is that the usual approach?

  19. Issy14
    • 2 years ago
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    am I correct on g(x)?

  20. geerky42
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    that's the problem and then they give you the graph

  23. geerky42
    • 2 years 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
    • 2 years ago
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    PS: \[\large\dfrac{d}{dx}|x| = \dfrac{x}{|x|}\]

  25. geerky42
    • 2 years 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
    • 2 years ago
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    what does "estimate the derivative" mean?

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

  28. geerky42
    • 2 years 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
    • 2 years ago
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    it's ok, I'm confused as well, thank you for your help.

  30. anonymous
    • 2 years ago
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    i think you are thinking too hard for this one

  31. anonymous
    • 2 years 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. anonymous
    • 2 years ago
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    similarly \(g(2)=2\) so \(g(g(2))=g(2)=2\)

  33. anonymous
    • 2 years ago
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    as for the derivative, \[\left(g(g(x)\right)'=g(g(x))g'(x)\] by the chain rule

  34. anonymous
    • 2 years ago
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    ok that was a mistake!!

  35. anonymous
    • 2 years ago
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    \[\left(g(g(x)\right)'=g'(g(x))g'(x)\]

  36. anonymous
    • 2 years ago
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    then \[w'(1)=g'(g(1))\times g'(1)=-1\times -1=1\]

  37. anonymous
    • 2 years ago
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    |dw:1404869501887:dw|

  38. anonymous
    • 2 years ago
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    which should not surprise you since \(g(g(x))=x\)

  39. Issy14
    • 2 years 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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