Results 1 to 2 of 2

Math Help - Finding the domain of an inverse function

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    12

    Finding the domain of an inverse function

    f(x)=e^{x^2}, x\geq0
    g(x)=\frac{1}{x+3}, x\neq-3

    (a) Find h(x) where h(x) = g \circ f(x).
    (b) State the domain of h^-1(x).
    (c) Find h^-1(x).

    The first question is easy to answer: h(x)=\frac{1}{e^{x^2}+3}. However, part (b) is what puzzles me. Without actually finding h^-1(x) (since that is asked for in (c)), I am supposed to discover the domain of it. In the markscheme the answer is simply stated, without explanation, and the question is not worth many marks. So it seems as if I am missing something very obvious here, doesn't it?

    According to the key, the answer is supposed to be 0 < x \leq \frac{1}{4}. How is this reached in a - supposedly - trivial way?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    The domain of the inverse is the image of the domain of the function h.

    Basically, take x≥0 and find what values you get for h.
    When x = 0, the denominator is 1/4.
    When x gets larger (i.e. approaches infinity), h goes to 1/infinity (0, in the extended sense).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 2nd 2012, 10:17 AM
  2. Replies: 1
    Last Post: April 11th 2010, 02:04 PM
  3. Finding the Domain of an Inverse Function
    Posted in the Calculus Forum
    Replies: 10
    Last Post: January 20th 2010, 08:47 PM
  4. Finding domain for inverse trig.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 15th 2009, 06:28 PM
  5. Finding the domain of an inverse sine function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 27th 2009, 08:30 PM

/mathhelpforum @mathhelpforum