Answer
1. \(\displaystyle (G \circ F)(x) = x.\)
2. \(\displaystyle (F \circ G)(x) = x.\)
3. They are **equal** and both are the identity function on \(\mathbb{R}\).
Work Step by Step
**Step 1. Identify the given functions.**
\(F(x) = x^5\).
\(G(x) = x^{1/5}\).
Both are defined for **all real** \(x\). (Because the 5th root function is well-defined for negative, zero, and positive inputs.)
---
## 1. Compute \(G \circ F\)
\[
(G \circ F)(x) \;=\; G\bigl(F(x)\bigr)
\;=\; G\bigl(x^5\bigr)
\;=\; \bigl(x^5\bigr)^{1/5}.
\] For real \(x\), since 5 is an **odd** exponent/root, \(\bigl(x^5\bigr)^{1/5} = x\). Thus,
\[
(G \circ F)(x) = x.
\]
## 2. Compute \(F \circ G\)
\[
(F \circ G)(x) \;=\; F\bigl(G(x)\bigr)
\;=\; F\bigl(x^{1/5}\bigr)
\;=\; \bigl(x^{1/5}\bigr)^5.
\]
Again, because the exponent and root are both 5 (an odd integer), \(\bigl(x^{1/5}\bigr)^5 = x\) for all real \(x\). Hence,
\[
(F \circ G)(x) = x.
\]
## 3. Are \(G \circ F\) and \(F \circ G\) Equal?
Yes. We have
\[
(G \circ F)(x) \;=\; x
\quad\text{and}\quad
(F \circ G)(x) \;=\; x,
\]
so **for all real \(x\)**,
\[
(G \circ F)(x) \;=\; (F \circ G)(x) \;=\; x.
\]
