We are tasked with finding the composite functions $(f \circ g)(x)$ and $(g \circ f)(x)$ for the given functions $f(x) = x^3 + 6$ and $g(x) = \sqrt[3]{x - 6}$.

1. Finding $(f \circ g)(x)$:

The composite function $f \circ g(x)$ means applying $g(x)$ first and then applying $f$ to the result.

• First, we have $g(x) = \sqrt[3]{x - 6}$.
• Now, substitute $g(x)$ into $f(x)$:

$(f \circ g)(x) = f(g(x)) = f\left(\sqrt[3]{x - 6}\right)$

Since $f(x) = x^3 + 6$, we substitute $\sqrt[3]{x - 6}$ for $x$ in $f(x)$:

$(f \circ g)(x) = \left(\sqrt[3]{x - 6}\right)^3 + 6$

Using the property that $(\sqrt[3]{a})^3 = a$, we simplify:

$(f \circ g)(x) = x - 6 + 6$

$(f \circ g)(x) = x$

2. Finding $(g \circ f)(x)$:

The composite function $g \circ f(x)$ means applying $f(x)$ first and then applying $g$ to the result.

• First, we have $f(x) = x^3 + 6$.
• Now, substitute $f(x)$ into $g(x)$:

$(g \circ f)(x) = g(f(x)) = g(x^3 + 6)$

Since $g(x) = \sqrt[3]{x - 6}$, we substitute $x^3 + 6$ for $x$ in $g(x)$:

$(g \circ f)(x) = \sqrt[3]{(x^3 + 6) - 6}$

Simplify the expression inside the cube root:

$(g \circ f)(x) = \sqrt[3]{x^3}$

Using the property that $\sqrt[3]{x^3} = x$, we simplify:

$(g \circ f)(x) = x$

Final Answers:

• $(f \circ g)(x) = x$
• $(g \circ f)(x) = x$

Both composite functions result in $x$.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. What is the general process for finding the composition of two functions?
2. How do the properties of cube roots and exponents help simplify composite functions?
3. What are the conditions for two functions to be inverses of each other?
4. How does the behavior of composite functions change with more complex functions?
5. Can composite functions have applications in real-world problems, such as physics or economics?

Tip: When dealing with composite functions, always start by carefully applying the inner function before substituting it into the outer function.