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Thread: Finding the domain of an inverse function

  1. #1
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    Finding the domain of an inverse function

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

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

    The first question is easy to answer: $\displaystyle h(x)=\frac{1}{e^{x^2}+3}$. However, part (b) is what puzzles me. Without actually finding $\displaystyle h^-1(x)$ (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 $\displaystyle 0 < x \leq \frac{1}{4}$. How is this reached in a - supposedly - trivial way?
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  2. #2
    Super Member TheChaz's Avatar
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    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).
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