Hmm, this is what I make of the question. (I mean, what Jamon said seems to make sense but I wanted to give a go myself anyway.)
1. \( f(x)\) is not invertible. It fails the horizontal line test since it involves the entire semi-circle.
2. \(g(x)\), however, is invertible. This falls out from the domain restriction.
Note that \(g(x)\) is the quadrant of the circle with domain \( 0\leq x \leq 1\) and range \(2 \leq x \leq 3\). The graph of the inverse can be, of course, determined by a simple reflection about \(y=x\), however at the very least we know that it has domain \(2 \leq x \leq 3\) and range \( 0\leq x \leq 1\).
3. I will temporarily overkill the question by also considering what happens when \( 0 \leq \phi \leq 1\). For such a fixed value of \(\phi\), we note that \(f(x)\) and \(g(x)\) represent the same function. Therefore, since \(g^{-1}(x)\) is the inverse function of \(g(x)\) over the relevant domains, it is also the inverse of \( f(x)\) here. Thus \(g^{-1}(f(x)) = x\) for all \(0\leq x \leq 1\). Therefore, in THIS case, \(g^{-1}(f(\phi)) = \phi\).
4. Now we consider what the question is asking for, i.e. when \( 1\leq \phi \leq 2\). Here, it is not immediately true that \( g^{-1} (f(\phi)) = \phi \) anymore. Reason being, \(\phi\) does not lie in the correct interval \(0 \leq x \leq 1\).
5. As it turns out, we can rewrite a (linear) expression
in terms of \(\phi\), that falls in the correct interval:
\[ 1\leq \phi \leq 2\\ -1\leq \phi-2\leq 0\\ 0\leq 2-\phi \leq 1 \]
So we see that \( \boxed{0\leq 2-\phi \leq 1} \). Therefore, we can deduce that \( \boxed{g^{-1}(f(2-\phi)) = 2-\phi }\).
6. Finally, we deal with the issue of symmetry. The above expression is still kinda useless, because we want \(\phi\) under the function \(f\), not \(2-\phi\). It turns out that it's a stepping stone.
\[ \text{The function }f(x) = 2+\sqrt{1-(x-1)^2}\text{, defined over the natural domain }0\leq x\leq 2\\ \text{is symmetric about }x=1\\ \text{thanks to the properties of the semi-circle.} \]
Picture
I mean, they give you a picture in the question, but here's one from desmos:
\[ \text{We know from our study of even functions in Year 11 that}\\ \text{if a curve is symmetric about the line }x=0\\ \text{then }f(x) = f(-x). \]
\[ \text{As it turns out, there is a theorem that states}\\ \text{if a curve is symmetric about the line}\boxed{x=a}\\ \text{then }\boxed{f(x)=f(2a-x)} \]
The proof of this theorem is typically done pictorially. It involves measuring the (perpendicular) distances of a curve that is symmetric about the line \(x=a\). Here's a rough idea: