# Thread: Finding the domain of an inverse function

1. ## 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?

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).