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Thread: f(x) = arctan(e^(-x)) + 1

  1. #1
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    f(x) = arctan(e^(-x)) + 1

    Let f(x)= arctan( e^-x ) + 1 ; x belongs to R

    1- Find range of f
    2- Determine whether is f 1-1 or not.
    3- Find inverse of f

    My answer:
    1- range of f is [1 , (2+pi)/pi ] .. I get it by evaluating limits of f at infinity and -infinity.
    2- f'(x) = - e^(-x) / ( e^(-2x) + 1 ) which is < 0 ----> f is decreasing ----> f is 1-1 ----> inverse exists.
    3- f^(-1)(x) = - ln ( tan (x-1) )

    Are my answers correct?
    Thanks
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  2. #2
    Member Walagaster's Avatar
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    Re: f(x) = arctan(e^(-x)) + 1

    Quote Originally Posted by TWiX View Post
    Let f(x)= arctan( e^-x ) + 1 ; x belongs to R

    1- Find range of f
    2- Determine whether is f 1-1 or not.
    3- Find inverse of f

    My answer:
    1- range of f is [1 , (2+pi)/pi ] .. I get it by evaluating limits of f at infinity and -infinity.
    2- f'(x) = - e^(-x) / ( e^(-2x) + 1 ) which is < 0 ----> f is decreasing ----> f is 1-1 ----> inverse exists.
    3- f^(-1)(x) = - ln ( tan (x-1) )

    Are my answers correct?
    Thanks
    They look good except I think you may have a typo on the second fraction in the range.
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  3. #3
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    Re: f(x) = arctan(e^(-x)) + 1

    ye it is supposed to be 2
    Thanks.
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