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Dido525

  • 3 years ago

Find a and b such that

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  1. Dido525
    • 3 years ago
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    the following function is differentiable for all values of x.

  2. Dido525
    • 3 years ago
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    |dw:1350070251929:dw|

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

  4. myininaya
    • 3 years ago
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    a=1 ?

  5. Dido525
    • 3 years ago
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    Oops. I figured out b = 1.

  6. Dido525
    • 3 years ago
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    I can`t solve for a.

  7. myininaya
    • 3 years ago
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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.

  8. myininaya
    • 3 years ago
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    Because one this it must need to be is continuous

  9. Dido525
    • 3 years ago
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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.

  10. Dido525
    • 3 years ago
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    Should I derive both functions and solve for a?

  11. Dido525
    • 3 years ago
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    But that just means it's continuous. Not differentiable.

  12. myininaya
    • 3 years ago
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    Right that is one condition we must need though

  13. myininaya
    • 3 years ago
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    We need it to run smoothly when they meet (no sharp corners)... hmm... thinking....

  14. Dido525
    • 3 years ago
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    So I go: |dw:1350070954216:dw|

  15. Dido525
    • 3 years ago
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    right? O_o .

  16. Dido525
    • 3 years ago
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    No. Thats not right.

  17. Dido525
    • 3 years ago
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    Never mind.

  18. myininaya
    • 3 years ago
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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

  19. Dido525
    • 3 years ago
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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.

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

  21. Dido525
    • 3 years ago
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    Right...

  22. Dido525
    • 3 years ago
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    I tried that.

  23. myininaya
    • 3 years ago
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    Find derivative of both sides Set x=0 And solve you are done

  24. Dido525
    • 3 years ago
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    Let me try again. Maybe I made a mistake.

  25. Dido525
    • 3 years ago
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    The derivative of arctan(ax+1) |dw:1350071299130:dw| right?

  26. myininaya
    • 3 years ago
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    Is that an a on top ?

  27. Dido525
    • 3 years ago
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    It shoud say 1/(1+(ax+1)^2)

  28. myininaya
    • 3 years ago
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    no it should say a on top

  29. myininaya
    • 3 years ago
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    that is your issue

  30. myininaya
    • 3 years ago
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    (ax+1)'=a

  31. Dido525
    • 3 years ago
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    Ohh. Why?

  32. Dido525
    • 3 years ago
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    Ohh....

  33. Dido525
    • 3 years ago
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    haha... Chain rule.... oops.

  34. myininaya
    • 3 years ago
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    \[(\arctan(u(x))'=\frac{u'(x)}{{1+(u(x))^2}}\]

  35. Dido525
    • 3 years ago
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    so it would be \[\frac{ a }{ 1+(ax+1)^2 }\]

  36. myininaya
    • 3 years ago
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    Yep and the other side you have?

  37. Dido525
    • 3 years ago
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    |dw:1350071580411:dw|

  38. Dido525
    • 3 years ago
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    right?

  39. Dido525
    • 3 years ago
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    I really appreciate the help by the way. Thanks a lot.

  40. myininaya
    • 3 years ago
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    I think you got it from here :) Great job.

  41. Dido525
    • 3 years ago
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    I got a = pi/2

  42. Dido525
    • 3 years ago
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    @myininaya .

  43. Dido525
    • 3 years ago
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    Thanks for making this so easy to understand! :) .

  44. myininaya
    • 3 years ago
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    That is what I got :)

  45. myininaya
    • 3 years ago
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    Gj again.

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