Answer
1. \(\displaystyle (g \circ f)(x) = -x - 3.\)
2. \(\displaystyle (g \circ f)^{-1}(x) = -x - 3.\)
3. \(\displaystyle g^{-1}(x) = -x.\)
4. \(\displaystyle f^{-1}(x) = x - 3.\)
5. \(\displaystyle (f^{-1}\circ g^{-1})(x) = -x - 3.\)
6. \(\displaystyle (g\circ f)^{-1} = f^{-1}\circ g^{-1}.\)
Work Step by Step
Below is a step‐by‐step solution. We have two functions:
\[
f(x) = x + 3
\quad\text{and}\quad
g(x) = -x.
\]
We wish to find:
1. \(g\circ f\)
2. \((g\circ f)^{-1}\)
3. \(g^{-1}\)
4. \(f^{-1}\)
5. \(f^{-1}\circ g^{-1}\)
6. Compare \((g\circ f)^{-1}\) and \(f^{-1}\circ g^{-1}\).
---
## 1) Compute \(g \circ f\)
\[
(g\circ f)(x)
\;=\;
g\bigl(f(x)\bigr)
\;=\;
g(x+3)
\;=\;
-(\,x+3\,)
\;=\;
-x - 3.
\]
Hence
\[
g \circ f : x \;\mapsto\; -x - 3.
\]
---
## 2) Find \((g \circ f)^{-1}\)
We have \((g\circ f)(x) = -x - 3\). To find the inverse, let
\[
y = -x - 3.
\]
Solve for \(x\):
\[
y + 3 = -x
\quad\Longrightarrow\quad
x = -\,y - 3.
\]
Therefore
\[
(g\circ f)^{-1}(y)
\;=\;
-\,y \;-\; 3.
\]
Renaming the dummy variable \(y\) back to \(x\), we get
\[
(g\circ f)^{-1}(x)
\;=\;
-\,x \;-\; 3.
\]
---
## 3) Find \(g^{-1}\)
Given \(g(x) = -x\), we check if \(g\) is its own inverse. Indeed, if \(y = -x\), then \(x = -y\). Thus
\[
g^{-1}(x) = -x.
\]
Hence \(g\) is an involution: \(g = g^{-1}\).
---
## 4) Find \(f^{-1}\)
Given \(f(x) = x+3\). If \(y = x+3\), then \(x = y-3\). Thus
\[
f^{-1}(x) = x - 3.
\]
---
## 5) Compute \(f^{-1}\circ g^{-1}\)
We apply \(g^{-1}\) first, then \(f^{-1}\). Since \(g^{-1}(x) = -x\),
\[
(f^{-1}\circ g^{-1})(x)
\;=\;
f^{-1}\bigl(g^{-1}(x)\bigr)
\;=\;
f^{-1}(-x)
\;=\;
(-x)\;-\;3
\;=\;
-\,x \;-\; 3.
\]
---
## 6) Relationship Between \((g \circ f)^{-1}\) and \(f^{-1}\circ g^{-1}\)
From Steps 2 and 5, both \((g \circ f)^{-1}\) and \(f^{-1}\circ g^{-1}\) map \(x\) to \(-x-3\). Therefore,
\[
\boxed{(g\circ f)^{-1} = f^{-1}\circ g^{-1}.}
\]
This matches the general fact that the inverse of a composition is the composition of the inverses in **reverse** order.
