To solve for the compositions of the functions $f(x) = x + 2$, $g(x) = x - 2$, and $h(x) = 3x$, let's work step by step through each composition. We'll use the definition of composition of functions: $(g \circ f)(x) = g(f(x))$, $(f \circ g)(x) = f(g(x))$, and so on.

1. Find $(g \circ f)(x) = g(f(x))$:

Substitute $f(x) = x + 2$ into $g(x) = x - 2$:

$g(f(x)) = g(x + 2)$

Since $g(x) = x - 2$, replace $x$ with $x + 2$:

$g(x + 2) = (x + 2) - 2 = x$

Thus, $(g \circ f)(x) = x$.

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2. Find $(f \circ g)(x) = f(g(x))$:

Substitute $g(x) = x - 2$ into $f(x) = x + 2$:

$f(g(x)) = f(x - 2)$

Since $f(x) = x + 2$, replace $x$ with $x - 2$:

$f(x - 2) = (x - 2) + 2 = x$

Thus, $(f \circ g)(x) = x$.

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3. Find $(f \circ f)(x) = f(f(x))$:

Substitute $f(x) = x + 2$ into itself:

$f(f(x)) = f(x + 2)$

Since $f(x) = x + 2$, replace $x$ with $x + 2$:

$f(x + 2) = (x + 2) + 2 = x + 4$

Thus, $(f \circ f)(x) = x + 4$.

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4. Find $(g \circ g)(x) = g(g(x))$:

Substitute $g(x) = x - 2$ into itself:

$g(g(x)) = g(x - 2)$

Since $g(x) = x - 2$, replace $x$ with $x - 2$:

$g(x - 2) = (x - 2) - 2 = x - 4$

Thus, $(g \circ g)(x) = x - 4$.

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Final Results:

1. $(g \circ f)(x) = x$
2. $(f \circ g)(x) = x$
3. $(f \circ f)(x) = x + 4$
4. $(g \circ g)(x) = x - 4$

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Would you like to explore these results further or analyze another example?

Related Questions:

1. What happens if we calculate $(h \circ f)$ or $(h \circ g)$?
2. What is the geometric interpretation of $(f \circ g)(x)$ and $(g \circ f)(x)$?
3. How do the compositions $f \circ f$ and $g \circ g$ behave for specific values of $x$?
4. What would happen if we introduced a new function $k(x) = x^2$?
5. How does function composition relate to inverses of functions?

Tip:

When working with compositions, always substitute step by step to avoid errors. Check whether the functions simplify to identities or create new transformations.