Answer
Both compositions give the identity on \(\mathbb{R}^+\). Therefore, \(G^{-1}\) is indeed the inverse of \(G\), and they satisfy
\[
G \circ G^{-1} = \mathrm{Id}_{\mathbb{R}^+}
\quad\text{and}\quad
G^{-1} \circ G = \mathrm{Id}_{\mathbb{R}^+}.
\]
Work Step by Step
We have the pair of functions
\[
G: \mathbb{R}^+ \to \mathbb{R}^+,
\quad G(x) = x^2,
\]
and its proposed inverse
\[
G^{-1} : \mathbb{R}^+ \to \mathbb{R}^+,
\quad G^{-1}(y) = \sqrt{y}.
\]
Here \(\mathbb{R}^+\) means the set of all positive real numbers. To verify \(G^{-1}\) is indeed the inverse of \(G\), we must show:
1. \(\bigl(G \circ G^{-1}\bigr)(y) = y\) for all \(y \in \mathbb{R}^+\).
2. \(\bigl(G^{-1} \circ G\bigr)(x) = x\) for all \(x \in \mathbb{R}^+\).
---
## 1. Compute \(\bigl(G \circ G^{-1}\bigr)(y)\)
\[
\bigl(G \circ G^{-1}\bigr)(y)
\;=\;
G\bigl(G^{-1}(y)\bigr)
\;=\;
G\bigl(\sqrt{y}\bigr)
\;=\;
(\sqrt{y})^2
\;=\;
y.
\]
This holds for all \(y>0\), so \(G\bigl(G^{-1}(y)\bigr) = y\).
---
## 2. Compute \(\bigl(G^{-1} \circ G\bigr)(x)\)
\[
\bigl(G^{-1} \circ G\bigr)(x)
\;=\;
G^{-1}\bigl(G(x)\bigr)
\;=\;
G^{-1}(x^2)
\;=\;
\sqrt{x^2}.
\]
Since \(x \in \mathbb{R}^+\), we have \(x>0\). Thus \(\sqrt{x^2} = x\). Hence
\[
G^{-1}\bigl(G(x)\bigr) = x
\]
for all \(x>0\).
