Answer: C) I, I and III
In other words, statements I through III are all true.
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Explanation:
Apply substitution to say that,
[tex]f(x) = 3x^3 + 2\\\\f(g(x)) = 3(g(x))^3 + 2\\\\f(g(x)) = 3\left(\sqrt[3]{\frac{x-2}{3}}\right)^3 + 2\\\\f(g(x)) = 3*\left(\frac{x-2}{3}\right) + 2\\\\f(g(x)) = x-2 + 2\\\\f(g(x)) = x\\\\[/tex]
We can see that the cube root and cube exponents cancel each other out, and everything simplifies to simply x.
Now compute the other composite function, where g(x) is the outer function this time.
[tex]g(x) = \sqrt[3]{\frac{x-2}{3}}\\\\g(f(x)) = \sqrt[3]{\frac{f(x)-2}{3}}\\\\g(f(x)) = \sqrt[3]{\frac{3x^3 + 2-2}{3}}\\\\g(f(x)) = \sqrt[3]{\frac{3x^3}{3}}\\\\g(f(x)) = \sqrt[3]{x^3}\\\\g(f(x)) = x\\\\[/tex]
Again, the cubing and cube root cancel each other out.
Since both f(g(x)) = x and g(f(x)) = x are true equations, this means that functions f(x) and g(x) are inverses of each other.
The two functions undo one another.
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As a simpler example, consider these two functions:
f(x) = x+5
g(x) = x-5
The f(x) function adds 5 to whatever the input is, and the g(x) goes in the opposite direction to subtract 5 to undo what f(x) did. For this example, you should find that f(g(x)) = x and g(f(x)) = x
Applying the composite of these inverses means that the final output is whatever the original input was. If we start with x = 87, then x+5 = 87+5 = 92, and then 92-5 = 87 which was the original input. Overall, no change has happened.
