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Dido525 Group Title

Find a and b such that

  • one year ago
  • one year ago

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  1. Dido525 Group Title
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    the following function is differentiable for all values of x.

    • one year ago
  2. Dido525 Group Title
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    |dw:1350070251929:dw|

    • one year ago
  3. Dido525 Group Title
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    I figured out a= 1 but I can't figure out what b is equal to.

    • one year ago
  4. myininaya Group Title
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    a=1 ?

    • one year ago
  5. Dido525 Group Title
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    Oops. I figured out b = 1.

    • one year ago
  6. Dido525 Group Title
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    I can`t solve for a.

    • one year ago
  7. myininaya Group Title
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    Yep. Now that is right so far. You just set the limx->0 for both of them and solve for a b. Good job.

    • one year ago
  8. myininaya Group Title
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    Because one this it must need to be is continuous

    • one year ago
  9. Dido525 Group Title
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    Yep . I set the both as lim x-> 0 and I got that b = 1. I can't solve for a however.

    • one year ago
  10. Dido525 Group Title
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    Should I derive both functions and solve for a?

    • one year ago
  11. Dido525 Group Title
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    But that just means it's continuous. Not differentiable.

    • one year ago
  12. myininaya Group Title
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    Right that is one condition we must need though

    • one year ago
  13. myininaya Group Title
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    We need it to run smoothly when they meet (no sharp corners)... hmm... thinking....

    • one year ago
  14. Dido525 Group Title
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    So I go: |dw:1350070954216:dw|

    • one year ago
  15. Dido525 Group Title
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    right? O_o .

    • one year ago
  16. Dido525 Group Title
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    No. Thats not right.

    • one year ago
  17. Dido525 Group Title
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    Never mind.

    • one year ago
  18. myininaya Group Title
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    Ok...how about this : What do you think of both parts of the function having the same exact slope at x=0 This will give us that smooth transition at x=0 that i'm talking about

    • one year ago
  19. Dido525 Group Title
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    Yep. I knew that. But unless I know what a is equal to I can't solve for the derivative at x=0.

    • one year ago
  20. myininaya Group Title
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    \[(\arctan(ax+1))'|_{x=0}=(\frac{\pi}{4} e^{\sin(x)})'|_{x=0}\]

    • one year ago
  21. Dido525 Group Title
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    Right...

    • one year ago
  22. Dido525 Group Title
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    I tried that.

    • one year ago
  23. myininaya Group Title
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    Find derivative of both sides Set x=0 And solve you are done

    • one year ago
  24. Dido525 Group Title
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    Let me try again. Maybe I made a mistake.

    • one year ago
  25. Dido525 Group Title
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    The derivative of arctan(ax+1) |dw:1350071299130:dw| right?

    • one year ago
  26. myininaya Group Title
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    Is that an a on top ?

    • one year ago
  27. Dido525 Group Title
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    It shoud say 1/(1+(ax+1)^2)

    • one year ago
  28. myininaya Group Title
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    no it should say a on top

    • one year ago
  29. myininaya Group Title
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    that is your issue

    • one year ago
  30. myininaya Group Title
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    (ax+1)'=a

    • one year ago
  31. Dido525 Group Title
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    Ohh. Why?

    • one year ago
  32. Dido525 Group Title
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    Ohh....

    • one year ago
  33. Dido525 Group Title
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    haha... Chain rule.... oops.

    • one year ago
  34. myininaya Group Title
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    \[(\arctan(u(x))'=\frac{u'(x)}{{1+(u(x))^2}}\]

    • one year ago
  35. Dido525 Group Title
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    so it would be \[\frac{ a }{ 1+(ax+1)^2 }\]

    • one year ago
  36. myininaya Group Title
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    Yep and the other side you have?

    • one year ago
  37. Dido525 Group Title
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    |dw:1350071580411:dw|

    • one year ago
  38. Dido525 Group Title
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    right?

    • one year ago
  39. Dido525 Group Title
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    I really appreciate the help by the way. Thanks a lot.

    • one year ago
  40. myininaya Group Title
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    I think you got it from here :) Great job.

    • one year ago
  41. Dido525 Group Title
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    I got a = pi/2

    • one year ago
  42. Dido525 Group Title
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    @myininaya .

    • one year ago
  43. Dido525 Group Title
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    Thanks for making this so easy to understand! :) .

    • one year ago
  44. myininaya Group Title
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    That is what I got :)

    • one year ago
  45. myininaya Group Title
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    Gj again.

    • one year ago
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