$\displaystyle f(x)=e^{x^2}, x\geq0$

$\displaystyle g(x)=\frac{1}{x+3}, x\neq-3$

(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?