Answer
1. \(\displaystyle (G \circ F)(x) = x^3 - 1.\)
2. \(\displaystyle (F \circ G)(x) = (x - 1)^3 = x^3 - 3x^2 + 3x - 1.\)
3. They are **not equal** as functions on \(\mathbb{R}\).
Work Step by Step
**Step 1. Identify the given functions.**
- \(F(x) = x^3\).
- \(G(x) = x - 1\).
Both are defined for all real \(x\).
---
## 1. Compute \(G \circ F\)
\[
(G \circ F)(x) \;=\; G\bigl(F(x)\bigr)
\;=\; G\bigl(x^3\bigr)
\;=\; x^3 \;-\; 1.
\]
---
## 2. Compute \(F \circ G\)
\[
(F \circ G)(x) \;=\; F\bigl(G(x)\bigr)
\;=\; F\bigl(x - 1\bigr)
\;=\; (\,x - 1\,)^3
\;=\; x^3 \;-\; 3x^2 \;+\; 3x \;-\; 1.
\]
---
## 3. Are \(G \circ F\) and \(F \circ G\) Equal?
- We have
\[
(G \circ F)(x) \;=\; x^3 - 1
\quad\text{and}\quad
(F \circ G)(x) \;=\; x^3 - 3x^2 + 3x - 1.
\]
- Compare the two expressions:
\[
x^3 - 1 \quad\text{vs.}\quad x^3 - 3x^2 + 3x - 1.
\]
Clearly, these are **not** the same polynomial unless \( -3x^2 + 3x = 0\), which only happens for specific \(x\) values (namely \(x=0\) or \(x=1\)).
Hence, for general \(x\in\mathbb{R}\),
\[
(G \circ F)(x) \;\neq\; (F \circ G)(x).
\]
