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Inverse functions, and limits help!

  1. Feb 10, 2009 #1
    Please help me with these following problems:

    1.)Indicate whether each of the following functions is invertible in the given interval. Explain
    a.) sech x on [0,infinity)

    b.) cos (ln x) on (O, e^pie]

    c.) e^(x^2) on (-1,2]


    2.) Evaluate the following limits, justifying your answers. If a limit does not exist explain why.

    a.) lim (x--> inf.) (3x^3 +cos x)/(sin x- x^3)
    b.) lim (x-->Pie(+)) (tan^-1 (1/(x-Pie)))/(Pie-x)
    c.) lim (x--> 0(+)) (sqrt(x+sin x))(ln x)
    d.) lim (x--> 1(-)) (cos^-1(x))/(1-x)

    3.) give the equation of the line tangent to the curve at the given point.
    a.) (y)(tan^-1 x)= x*y at (sqrt(3),0)
    b.) ln y= x^2 +(2)*e^x at (0, e^2)


    My attempt at a solution for a.)

    1.) sech x on [0,infinity)

    let y= sech x
    therefore, y*sech^-1 x= sech^-1 x* sech x

    therefore x= y*sech^-1 x

    since it is an inverse function switch x and y variables

    therefore f^-1(x)= x*sech^-1 (y)

    how do i determine, it's intervible in [0, infinity)... i'm not sure if this prior work is correct
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2009 #2

    Tom Mattson

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    That won't do. You've simply assumed the existence of sech^-1(x). You need to show that the function is 1-1 on [itex][0,\infty)[/itex] (that is, you need to show that it passes the so-called horizontal line test on that interval). One way to do that is to show that the function is monotonic on that interval. Have you been taught a theorem that shows you how to determine the monotonicity of a function?
     
  4. Feb 10, 2009 #3
    yes, basically for the horizontal line test, each x values only one unique f(x) value if it is to be one-to-one...

    since, sech(x) = 1 / cosh(x) = 2 / (e^x + e^-x)

    can i compute y= 2/(e^x +e^-x)

    and solve for x:

    than switch the variables for y and x to find the inverse function?

    and then plug in the values of [0,infinity) into x

    and determine if a f(x) value exists?

    is this a proper method
     
  5. Feb 10, 2009 #4

    Tom Mattson

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    In principle that's fine, but in practice it's a pain. It would be easier to show that y=sech(x) is monotonic (that is, either always increasing or always decreasing). Now the question for you is, how do you tell whether a function is increasing or decreasing on an interval?
     
  6. Feb 10, 2009 #5
    but wouldn't I have to determine the graph of the inverse of the sech (x) equation in order to determine if a value exists for [0, infinity)
     
  7. Feb 11, 2009 #6

    Tom Mattson

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    You don't have to find the inverse, you just have to determine whether or not an inverse exists. There's a big difference. You need to use the following facts.

    1.) A function [itex]f:D\longrightarrow R[/itex] has an inverse [itex]f^{-1}:R\longrightarrow D[/itex] if and only if [itex]f[/itex] is 1-1 and onto. The onto part is easy: just assume that the output values of the function are the range of the function. It's the 1-1 part that you need to worry about. This brings is to...

    2.) A function is 1-1 on its domain if and only if it is monotonic on its domain.

    So how do you show that [itex]f[/itex] is monotonic?
     
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