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Inverse Equation and if Values exist!

  1. Feb 11, 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]

    My work process for a: let y= sech x

    how do i find x in terms of y for this eq. how do i isolate the x variable from sech x.

    once I find that i can see if the positive values of x for the inverse function exist from 0 to infinity.

    however, i need help finding the inverse function

    My work process for b.) let y= cos (ln x)

    I took the expotential of both sides to get

    e^y= e^(cos*(lnx))

    how do i isoloate e*(lnx) to equal x , since the lnx is a part of the cosine angle expression.. i need help here...

    once i find the inverse equation, i can see if y values exists for the x values from 0 to e^pie... i need help finding the inverse equation first

    My work process for c.) let = y= e^(x^2)

    therefore taking the natural logarithm of both sides i get

    ln y= ln (e^(x^2)
    ln y= x^2

    therefore x= sqrt (ln y)

    therefore f^-1 (x)---> inverse function is f(x)= sqrt (ln x)

    when subbing (-1,2] for x to determine if y values exist,

    i find the inverse function doesn't exist under those limits since ln 0 is undefined

    is this the correct approach to this problem.

    Please help me with these problems!
  2. jcsd
  3. Feb 11, 2009 #2


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    Science Advisor

    A function is invertible in an interval if it is "one-to-one" there: that is, that two different values of x in the interval that give the same value of y. It should be fairly obvious, for example, that (-.5)2= .52 so [itex]e^{(-.5)^2}= e^(0.5)^2[/itex].

    You say at one point
    which is not true: if ln y= x^2 then either x= sqrt(ln y) or x= -sqrt(ln y).

    Your point about ln(0) not existing is a good one.

    For the others try looking at graphs of the functions.
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