To prove that two functions are inverses of each other, we need to show that composing one function with the other yields the identity function. Specifically, [tex]\( f(g(x)) \)[/tex] should equal [tex]\( x \)[/tex] and [tex]\( g(f(x)) \)[/tex] should equal [tex]\( x \)[/tex].
Let's start with the compositions:
### 1. Proving [tex]\( f(g(x)) = x \)[/tex]
Given:
- [tex]\( f(x) = 3(x + 2)^3 - 1 \)[/tex]
- [tex]\( g(x) = \sqrt[3]{\frac{x + 1}{3}} - 2 \)[/tex]
First, compute [tex]\( f(g(x)) \)[/tex]:
[tex]\[
g(x) = \sqrt[3]{\frac{x + 1}{3}} - 2
\][/tex]
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(g(x)) = f\left( \sqrt[3]{\frac{x+1}{3}} - 2 \right)
\][/tex]
Now, apply the function [tex]\( f \)[/tex]:
[tex]\[
f\left( \sqrt[3]{\frac{x+1}{3}} - 2 \right) = 3 \left( \left( \sqrt[3]{\frac{x+1}{3}} - 2 \right) + 2 \right)^3 - 1
\][/tex]
Simplify inside the parentheses:
[tex]\[
= 3 \left( \sqrt[3]{\frac{x+1}{3}} \right)^3 - 1
\][/tex]
Since [tex]\( (\sqrt[3]{a})^3 = a \)[/tex]:
[tex]\[
= 3 \left( \frac{x+1}{3} \right) - 1
\][/tex]
[tex]\[
= (x + 1) - 1
\][/tex]
[tex]\[
= x
\][/tex]
So, we have shown that [tex]\( f(g(x)) = x \)[/tex].
### 2. Proving [tex]\( g(f(x)) = x \)[/tex]
Now, compute [tex]\( g(f(x)) \)[/tex]:
Given:
- [tex]\( f(x) = 3(x + 2)^3 - 1 \)[/tex]
- [tex]\( g(x) = \sqrt[3]{\frac{x + 1}{3}} - 2 \)[/tex]
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[
g(f(x)) = g\left( 3(x + 2)^3 - 1 \right)
\][/tex]
Now, apply the function [tex]\( g \)[/tex]:
[tex]\[
g\left( 3(x+2)^3 - 1 \right) = \sqrt[3]{\frac{3(x+2)^3 - 1 + 1}{3}} - 2
\][/tex]
Simplify inside the cube root:
[tex]\[
= \sqrt[3]{\frac{3(x+2)^3}{3}} - 2
\][/tex]
[tex]\[
= \sqrt[3]{(x + 2)^3} - 2
\][/tex]
Since [tex]\( \sqrt[3]{a^3} = a \)[/tex]:
[tex]\[
= x + 2 - 2
\][/tex]
[tex]\[
= x
\][/tex]
So, we have shown that [tex]\( g(f(x)) = x \)[/tex].
Thus, since both [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex], we have proven that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are indeed inverses of each other.
