Answer :
To prove that two functions are inverses of each other, you need to show that the compositions of the functions [tex]\( f(g(x)) \)[/tex] and [tex]\( g(f(x)) \)[/tex] both simplify to [tex]\( x \)[/tex]. This means that when you apply one function to the result of the other function, you end up with [tex]\( x \)[/tex], which is the original input value.
Let's begin with the functions [tex]\( f(x) = 3(x+2)^3 - 1 \)[/tex] and [tex]\( g(x) = \sqrt[3]{\frac{x+1}{3}} - 2 \)[/tex]. We will show algebraically that [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex].
### Proof:
1. Composition [tex]\( f(g(x)) \)[/tex]:
Start with [tex]\( g(x) = \sqrt[3]{\frac{x+1}{3}} - 2 \)[/tex] and substitute it into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\sqrt[3]{\frac{x+1}{3}} - 2\right) \][/tex]
Now, substitute [tex]\( \sqrt[3]{\frac{x+1}{3}} - 2 \)[/tex] for [tex]\( x \)[/tex] in [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]\( \left(\sqrt[3]{\frac{x+1}{3}}\right)^3 = \frac{x+1}{3} \)[/tex]:
[tex]\[ 3\left(\frac{x+1}{3}\right) - 1 \][/tex]
Now, multiply by 3:
[tex]\[ x + 1 - 1 \][/tex]
Simplify:
[tex]\[ x \][/tex]
Thus,
[tex]\[ f(g(x)) = x \][/tex]
2. Composition [tex]\( g(f(x)) \)[/tex]:
Start with [tex]\( f(x) = 3(x+2)^3 - 1 \)[/tex] and substitute it into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(3(x+2)^3 - 1) \][/tex]
Now, substitute [tex]\( 3(x+2)^3 - 1 \)[/tex] for [tex]\( x \)[/tex] in [tex]\( g \)[/tex]:
[tex]\[ g(3(x+2)^3 - 1) = \sqrt[3]{\frac{(3(x+2)^3 - 1) + 1}{3}} - 2 \][/tex]
Simplify inside the fraction:
[tex]\[ \sqrt[3]{\frac{3(x+2)^3 - 1 + 1}{3}} - 2 \][/tex]
This simplifies further to:
[tex]\[ \sqrt[3]{\frac{3(x+2)^3}{3}} - 2 \][/tex]
Since [tex]\( \frac{3(x+2)^3}{3} = (x+2)^3 \)[/tex]:
[tex]\[ \sqrt[3]{(x+2)^3} - 2 \][/tex]
Which simplifies to:
[tex]\[ x + 2 - 2 \][/tex]
Simplify:
[tex]\[ x \][/tex]
Thus,
[tex]\[ g(f(x)) = x \][/tex]
Since both compositions [tex]\( f(g(x)) \)[/tex] and [tex]\( g(f(x)) \)[/tex] simplify to [tex]\( x \)[/tex], the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are indeed inverses of each other.
Let's begin with the functions [tex]\( f(x) = 3(x+2)^3 - 1 \)[/tex] and [tex]\( g(x) = \sqrt[3]{\frac{x+1}{3}} - 2 \)[/tex]. We will show algebraically that [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex].
### Proof:
1. Composition [tex]\( f(g(x)) \)[/tex]:
Start with [tex]\( g(x) = \sqrt[3]{\frac{x+1}{3}} - 2 \)[/tex] and substitute it into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\sqrt[3]{\frac{x+1}{3}} - 2\right) \][/tex]
Now, substitute [tex]\( \sqrt[3]{\frac{x+1}{3}} - 2 \)[/tex] for [tex]\( x \)[/tex] in [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]\( \left(\sqrt[3]{\frac{x+1}{3}}\right)^3 = \frac{x+1}{3} \)[/tex]:
[tex]\[ 3\left(\frac{x+1}{3}\right) - 1 \][/tex]
Now, multiply by 3:
[tex]\[ x + 1 - 1 \][/tex]
Simplify:
[tex]\[ x \][/tex]
Thus,
[tex]\[ f(g(x)) = x \][/tex]
2. Composition [tex]\( g(f(x)) \)[/tex]:
Start with [tex]\( f(x) = 3(x+2)^3 - 1 \)[/tex] and substitute it into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(3(x+2)^3 - 1) \][/tex]
Now, substitute [tex]\( 3(x+2)^3 - 1 \)[/tex] for [tex]\( x \)[/tex] in [tex]\( g \)[/tex]:
[tex]\[ g(3(x+2)^3 - 1) = \sqrt[3]{\frac{(3(x+2)^3 - 1) + 1}{3}} - 2 \][/tex]
Simplify inside the fraction:
[tex]\[ \sqrt[3]{\frac{3(x+2)^3 - 1 + 1}{3}} - 2 \][/tex]
This simplifies further to:
[tex]\[ \sqrt[3]{\frac{3(x+2)^3}{3}} - 2 \][/tex]
Since [tex]\( \frac{3(x+2)^3}{3} = (x+2)^3 \)[/tex]:
[tex]\[ \sqrt[3]{(x+2)^3} - 2 \][/tex]
Which simplifies to:
[tex]\[ x + 2 - 2 \][/tex]
Simplify:
[tex]\[ x \][/tex]
Thus,
[tex]\[ g(f(x)) = x \][/tex]
Since both compositions [tex]\( f(g(x)) \)[/tex] and [tex]\( g(f(x)) \)[/tex] simplify to [tex]\( x \)[/tex], the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are indeed inverses of each other.
