To solve the problem $(f \circ g) \circ h = (f \circ g \circ h)(x)$, let's break down the operations using the definitions of function composition.

Step-by-step Explanation

1. Understand the Notation:

   • $f \circ g$ denotes the composition of functions $f$ and $g$, which means $f(g(x))$.
   • Similarly, $(f \circ g) \circ h$ denotes the composition of the function $f \circ g$ with $h$.

2. Composition of Functions:

   • $(f \circ g) \circ h$ means applying $h$ first, then $g$, and finally $f$.
   • In terms of function notation, $(f \circ g) \circ h (x) = (f \circ g)(h(x))$.

3. Break Down the Composition:

   • Start with the inner function and apply each function step-by-step: $(f \circ g) \circ h (x) = f(g(h(x)))$ This means we:
   • First, apply $h$ to $x$: $h(x)$.
   • Then apply $g$ to the result: $g(h(x))$.
   • Finally, apply $f$ to this result: $f(g(h(x)))$.

4. Compare with $(f \circ g \circ h)(x)$:

   • By definition, $f \circ g \circ h$ means apply $h$, then $g$, and finally $f$, which is: $(f \circ g \circ h)(x) = f(g(h(x))).$

5. Conclusion:

   • Both expressions, $(f \circ g) \circ h(x)$ and $(f \circ g \circ h)(x)$, result in $f(g(h(x)))$.

Thus, $(f \circ g) \circ h = f \circ g \circ h$ when applied to any input $x$.

Summary

The compositions $(f \circ g) \circ h$ and $f \circ g \circ h$ are equivalent in this context, meaning their outputs are the same for all $x$.

Would you like more details on function composition or have any questions?

Here are 5 related questions for further exploration:

1. What is the composition of three different functions $f(x) = x + 1$, $g(x) = 2x$, and $h(x) = x^2$?
2. How do you find the inverse of a composed function like $f(g(x))$?
3. Can function composition be associative and commutative? Why or why not?
4. What happens if one of the functions in the composition is not defined for certain values of $x$?
5. How do you differentiate a composition of functions, such as $f(g(h(x)))$?

Tip: Remember, function composition follows the order from right to left when applying functions, so always start with the innermost function!