To prove that the two functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses of each other, we need to show that the composition of one function inside the other returns the original input value: [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex].
### Explanation (4 points):
Proving that two functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are inverses involves demonstrating that composing [tex]\( f \)[/tex] with [tex]\( g \)[/tex] (and vice versa) results in the identity function. Specifically, we need to show:
1. [tex]\( f(g(x)) = x \)[/tex]
2. [tex]\( g(f(x)) = x \)[/tex]
### Algebraic Proof (6 points):
#### 1. Calculate [tex]\( f(g(x)) \)[/tex]:
Given [tex]\( f(x) = 3(x + 2)^3 - 1 \)[/tex] and [tex]\( g(x) = \sqrt[3]{\frac{x + 1}{3}} - 2 \)[/tex]:
First, we compose [tex]\( f \)[/tex] with [tex]\( g \)[/tex]:
[tex]\[ f(g(x)) = 3 \left( \left(\sqrt[3]{\frac{x + 1}{3}} - 2 \right) + 2 \right)^3 - 1 \][/tex]
Simplify inside the parentheses:
[tex]\[ f(g(x)) = 3 \left( \sqrt[3]{\frac{x + 1}{3}} \right)^3 - 1 \][/tex]
Since [tex]\(\left( \sqrt[3]{\frac{x + 1}{3}} \right)^3 = \frac{x + 1}{3}\)[/tex], we get:
[tex]\[ f(g(x)) = 3 \left( \frac{x + 1}{3} \right) - 1 \][/tex]
[tex]\[ f(g(x)) = x + 1 - 1 \][/tex]
[tex]\[ f(g(x)) = x \][/tex]
#### 2. Calculate [tex]\( g(f(x)) \)[/tex]:
Given [tex]\( g(x) = \sqrt[3]{\frac{x + 1}{3}} - 2 \)[/tex] and [tex]\( f(x) = 3(x + 2)^3 - 1 \)[/tex]:
Next, we compose [tex]\( g \)[/tex] with [tex]\( f \)[/tex]:
[tex]\[ g(f(x)) = \sqrt[3]{\frac{3(x + 2)^3 - 1 + 1}{3}} - 2 \][/tex]
Simplify inside the fraction:
[tex]\[ g(f(x)) = \sqrt[3]{\frac{3(x + 2)^3}{3}} - 2 \][/tex]
[tex]\[ g(f(x)) = \sqrt[3]{(x + 2)^3} - 2 \][/tex]
Since [tex]\(\sqrt[3]{(x + 2)^3} = x + 2\)[/tex], we get:
[tex]\[ g(f(x)) = (x + 2) - 2 \][/tex]
[tex]\[ g(f(x)) = x \][/tex]
Thus, we have shown that:
[tex]\[ f(g(x)) = x \][/tex]
[tex]\[ g(f(x)) = x \][/tex]
This confirms that the functions [tex]\( f(x) = 3(x + 2)^3 - 1 \)[/tex] and [tex]\( g(x) = \sqrt[3]{\frac{x + 1}{3}} - 2 \)[/tex] are indeed inverses of each other.
