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Issy14

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

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

  3. Issy14
    • 5 months ago
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    |dw:1404857483770:dw|

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

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

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

  10. Issy14
    • 5 months ago
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    help @jdoe0001

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

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

  13. Issy14
    • 5 months ago
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    with the problem given

  14. Issy14
    • 5 months ago
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    it comes with the graph and the instructions.

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

  16. Issy14
    • 5 months ago
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    @geerky42

  17. geerky42
    • 5 months ago
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    yeah, that's how I figure it out

  18. Issy14
    • 5 months ago
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    alright, is that the usual approach?

  19. Issy14
    • 5 months ago
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    am I correct on g(x)?

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

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

  25. geerky42
    • 5 months 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
    • 5 months ago
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    what does "estimate the derivative" mean?

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

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

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

  31. satellite73
    • 5 months 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
    • 5 months ago
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    similarly \(g(2)=2\) so \(g(g(2))=g(2)=2\)

  33. satellite73
    • 5 months 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
    • 5 months ago
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    ok that was a mistake!!

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

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

  37. febylailani
    • 5 months ago
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    |dw:1404869501887:dw|

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

  39. Issy14
    • 5 months 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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