<Question>
Let $f(x)$ be an odd function, and let $g(x)$ be an even function.  Is $f(f(g(f(g(f(x))))))$ even, odd, or neither?

Enter "odd", "even", or "neither".
</Question>

<think>
<Plan_1>
Okay, let's try to figure out whether the composite function f(f(g(f(g(f(x)) ) ) ) is even, odd, or neither. Hmm, so we have f as an odd function and g as an even function. I remember that odd functions satisfy the property h(-x) = -h(x) and even functions satisfy h(-x) = h(x).
Maybe I should start by recalling the definitions and then work through the composition step by step.

First, let's recall the definitions:

<prompt_1.1> Recall the definitions of even and odd functions and consider how they apply to the given composition of functions. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
An even function g(x) satisfies g(-x) = g(x) for all x. An odd function f(x) satisfies f(-x) = -f(x) for all x. Now, the composite function in question is f(f(g(f(g(f(x)) )) ) ). Let's break this down layer by layer. Maybe starting from the innermost function and working our way out.

Let's denote each layer for clarity. Let's define:

h₁(x) = f(x)

h₂(x) = g(h₁(x)) = g(f(x))

h₃(x) = f(h₂(x)) = f(g(f(x)))

h₄(x) = g(h₃(x)) = g(f(g(f(x))))

h₅(x) = f(h₄(x)) = f(g(f(g(f(x)) )) )

h₆(x) = g(h₅(x)) = g(f(g(f(g(f(x)) )) )) 

Finally, the composite function is h₆(x) = f(f(g(f(g(f(x)) )) ) ).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but maybe I need to check if the entire composite function is even, odd, or neither. To do that, I need to evaluate h₆(-x) and see if it equals h₆(x) (even), -h₆(x) (odd), or neither.

Let's compute h₆(-x). Let's proceed step by step.

<prompt_2.1> Evaluate the composite function at -x and determine if it matches the original function or its negative. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Starting with h₁(-x) = f(-x). Since f is odd, f(-x) = -f(x). So h₁(-x) = -h₁(x).

Then h₂(-x) = g(h₁(-x)) = g(-h₁(x)). But g is even, so g(-h₁(x)) = g(h₁(x)). Therefore, h₂(-x) = h₂(x). So h₂ is even.

Next, h₃(-x) = f(h₂(-x)) = f(h₂(x)). Since h₂ is even, h₂(-x) = h₂(x). So h₃(-x) = f(h₂(x)) = h₃(x). Wait, is that right? Wait, h₃(x) = f(h₂(x)), so h₃(-x) = f(h₂(-x)) = f(h₂(x)) = h₃(x). Therefore, h₃ is even?

Wait, but f is odd. Let me verify. Suppose h₂ is even, so h₂(-x) = h₂(x). Then h₃(-x) = f(h₂(-x)) = f(h₂(x)) = h₃(x). So yes, h₃ is even. Hmm.

Then h₄(-x) = g(h₃(-x)) = g(h₃(x)). Since g is even, this is g(h₃(x)) = h₄(x). Therefore, h₄ is even.

Then h₅(-x) = f(h₄(-x)) = f(h₄(x)). Since h₄ is even, h₄(-x) = h₄(x). So h₅(-x) = f(h₄(x)) = h₅(x). Therefore, h₅ is even.

Then h₆(-x) = g(h₅(-x)) = g(h₅(x)). Since g is even, this is g(h₅(x)) = h₆(x). Therefore, h₆ is even.

Wait, so according to this step-by-step, each subsequent function alternates between even and odd based on the previous function.
Starting from h₁ (odd), h₂ (even), h₃ (even), h₄ (even), h₅ (even), h₆ (even). So the final function h₆ is even.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But let me double-check because sometimes when you have multiple compositions, the parity can change in a different way. Let me verify each step again.

<prompt_3.1> Re-evaluate each step of the composition to verify the parity of each function. </prompt_3.1>

But wait, let's test with a concrete example to make sure. Let's take f(x) = x (which is odd) and g(x) = x² (which is even).

<prompt_3.2> Test the conclusion with specific examples of odd and even functions. </prompt_3.2>

But let me try another example to confirm. Let f(x) = x^3 (odd) and g(x) = cos(x) (even).

<prompt_3.3> Test the conclusion with another set of examples to confirm the result. </prompt_3.3>

But let me check another example where the functions are different. Let f(x) = sin(x) (odd) and g(x) = x^2 (even).

<prompt_3.4> Test the conclusion with additional examples to ensure consistency. </prompt_3.4>

But to be thorough, let me check with a more complex example. Let f(x) = e^x - e^{-x} (which is odd because f(-x) = e^{-x} - e^x = -f(x)) and g(x) = x^4 (even).

<prompt_3.5> Test the conclusion with a more complex example to ensure the final answer is correct. </prompt_3.5>
</Plan_3>

<Execution_3>
<execution_3.1>
1. h₁(x) = f(x) (odd)
2. h₂(x) = g(f(x)) (even because g is even and composition of even with odd is even? Wait, hold on. Let me think. If g is even, then g(-y) = g(y). So if we input -y into g, it's the same as y. So if h₂(x) = g(f(x)), then h₂(-x) = g(f(-x)) = g(-f(x)) because f is odd. Then since g is even, g(-f(x)) = g(f(x)). Therefore, h₂(-x) = g(f(x)) = h₂(x). So yes, h₂ is even.
3. h₃(x) = f(g(f(x))). Let's check h₃(-x) = f(g(f(-x))) = f(g(-f(x))) because f is odd. Then since g is even, g(-f(x)) = g(f(x)). Therefore, h₃(-x) = f(g(f(x))) = h₃(x). So h₃ is even.
4. h₄(x) = g(f(g(f(x)))). Then h₄(-x) = g(f(g(f(-x)))) = g(f(g(-f(x)))) because f is odd. Then g(-f(x)) = g(f(x)) since g is even. So inside becomes g(f(x)), then f(g(f(x))) is h₃(x), which is even. Wait, no. Wait, h₄(-x) = g(f(g(f(-x)))) = g(f(g(-f(x)))) = g(f(g(f(x)))) because g(-f(x)) = g(f(x)). Therefore, h₄(-x) = h₄(x). So h₄ is even.
5. h₅(x) = f(g(f(g(f(x)) ))). Then h₅(-x) = f(g(f(g(f(-x)) ))). Let's compute the inside: f(-x) = -f(x), so g(f(-x)) = g(-f(x)) = g(f(x)) because g is even. Then f(g(f(-x))) = f(g(f(x))) which is h₃(x). Then h₃(x) is even, so f(h₃(x)) = h₅(x). Wait, but h₅(-x) = f(g(f(g(f(-x)) )) ) = f(g(f(g(-f(x)) )) ). Then g(-f(x)) = g(f(x)), so this becomes f(g(f(g(f(x)) )) ) = h₅(x). Therefore, h₅ is even.
6. h₆(x) = g(f(g(f(g(f(x)) )) )). Then h₆(-x) = g(f(g(f(g(f(-x)) )) )). Let's compute step by step:

f(-x) = -f(x)

g(f(-x)) = g(-f(x)) = g(f(x)) because g is even.

f(g(f(-x))) = f(g(f(x))) which is h₃(x). Then h₃(-x) = h₃(x) because h₃ is even. So f(h₃(x)) = h₅(x). Then h₅(-x) = h₅(x). Then g(h₅(-x)) = g(h₅(x)) = h₆(x). Therefore, h₆(-x) = h₆(x). So h₆ is even.
</execution_3.1>

<execution_3.2>
Compute f(f(g(f(g(f(x)))) )).

Start with the innermost function:

f(x) = x.

g(f(x)) = g(x) = x².

f(g(f(x))) = f(x²) = (x²)^3 = x^6.

g(f(g(f(x)))) = g(x^6) = (x^6)^2 = x^{12}.

f(f(g(f(g(f(x)) )) )) = f(x^{12}) = (x^{12})^3 = x^{36}.

Now, check if x^{36} is even, odd, or neither. Since x^{36} is x raised to an even power, it's an even function. So in this case, the composite function is even.
</execution_3.2>

<execution_3.3>
Compute the composition:

f(f(g(f(g(f(x)) )) )) = f(f(g(f(g(x³)) )) ) 

First, compute innermost: f(x) = x³. Then g(f(x)) = cos(x³). Then f(g(f(x))) = f(cos(x³)) = (cos(x³))³. Then g(f(g(f(x)))) = g((cos(x³))³) = [(cos(x³))³]^2 = cos²(x³)^3. Wait, no. Wait, g(x) = cos(x), so g((cos(x³))³) = cos( (cos(x³))³ ). Then f(g(f(g(f(x)) )) ) = [cos( (cos(x³))³ )]^3. Then the outermost f is applied again: f(f(...)) = [ [cos( (cos(x³))³ )]^3 ]^3 = [cos( (cos(x³))³ )]^9. Now, is this function even or odd? Let's check h(-x). Let's replace x with -x:

[cos( (cos( (-x)³ ))³ )]^9. Since (-x)³ = -x³, so cos(-x³) = cos(x³). Therefore, inside the cosine, we have cos(-x³) = cos(x³). Then we have [cos( (cos(x³))³ )]^9, which is the same as h(x). Therefore, h(-x) = h(x), so it's even.
</execution_3.3>

<execution_3.4>
Compute the composition:

f(f(g(f(g(f(x)) )) )) 

First, compute f(x) = x^3. Then g(f(x)) = (x^3)^2 = x^6. Then f(g(f(x))) = (x^6)^3 = x^{18}. Then g(f(g(f(x)))) = (x^{18})^2 = x^{36}. Then f(g(f(g(f(x)) )) ) = (x^{36})^3 = x^{108}. Then f(f(g(f(g(f(x)) )) )) = (x^{108})^3 = x^{324}. Now, h₆(x) = x^{324}. Then h₆(-x) = (-x)^{324} = (-1)^{324} x^{324} = x^{324} since 324 is even. Therefore, h₆(-x) = h₆(x), so it's even. Which matches our previous conclusion.

Another example: Let f(x) = x (odd) and g(x) = |x| (even). Then compute the composition:

f(f(g(f(g(f(x)) )) )) 

Start with f(x) = x. Then g(f(x)) = |x|. Then f(g(f(x))) = ||x|| = |x|. Then g(f(g(f(x)))) = ||x|| = |x|. Then f(g(f(g(f(x)) )) ) = ||x|| = |x|. Then f(f(g(f(g(f(x)) )) )) = ||x|| = |x|. So again, h₆(x) = |x|, which is even. So again, even.
</execution_3.4>

<execution_3.5>
Then let's compute the composition step by step:

First, f(x) = e^x - e^{-x}

g(x) = x^4

Compute f(g(f(g(f(x)) )) )

First, compute f(x): e^x - e^{-x}

Then g(f(x)) = (e^x - e^{-x})^4

Then f(g(f(x))) = f( (e^x - e^{-x})^4 ) = e^{(e^x - e^{-x})^4} - e^{ - (e^x - e^{-x})^4 }

Then g(f(g(f(x)))) = [ e^{(e^x - e^{-x})^4} - e^{ - (e^x - e^{-x})^4 } ]^4

Then f(g(f(g(f(x)) )) ) = f( [ e^{(e^x - e^{-x})^4} - e^{ - (e^x - e^{-x})^4 } ]^4 )

Now, check if this is even or odd. Let's substitute -x:

f(g(f(g(f(-x)) )) ) 

First, f(-x) = -f(x) = - (e^x - e^{-x})

Then g(f(-x)) = g(-f(x)) = (-f(x))^4 = (f(x))^4 = g(f(x)) since g is even.

Then f(g(f(-x))) = f(g(f(x))) as before.

Then g(f(g(f(-x)))) = g(f(g(f(x)))) same as before.

Then f(g(f(g(f(-x)) )) ) = f(g(f(g(f(x)) )) ) with an odd function applied to an even function's output.

Wait, but in this case, the outermost function is f, which is odd. The argument to f is g(f(g(f(-x)))) which is g(f(g(f(x)))) because g is even. So the argument is the same, so f( same ) = -f( original argument ). Wait, but hold on.

Wait, let's step through it:

Let’s denote the inner function as follows:

Let’s let’s define:

Let’s let’s define:

Let’s define the innermost function as A(x) = f(x)

Then B(x) = g(A(x))

Then C(x) = f(B(x))

Then D(x) = g(C(x))

Then E(x) = f(D(x))

So the composite function is E(x) = f(g(f(g(f(x)) )) )

Now, compute E(-x):

E(-x) = f(D(-x)) 

But D(x) = g(C(x)), so D(-x) = g(C(-x))

C(x) = f(B(x)), so C(-x) = f(B(-x))

B(x) = g(A(x)), so B(-x) = g(A(-x))

A(x) = f(x), so A(-x) = f(-x) = -f(x) (since f is odd)

Therefore, B(-x) = g(-f(x)). Since g is even, g(-f(x)) = g(f(x)) = B(x). So B(-x) = B(x)

Therefore, C(-x) = f(B(-x)) = f(B(x)) = C(x). So C(-x) = C(x)

Then D(-x) = g(C(-x)) = g(C(x)) = D(x). So D(-x) = D(x)

Then E(-x) = f(D(-x)) = f(D(x)) = E(x). Therefore, E(-x) = E(x), so E is even.

Therefore, regardless of the functions (as long as f is odd and g is even), the composition ends up being even.
</execution_3.5>
</Execution_3>

<Plan_4>
Based on execution_3.5:
**Final Answer**
\boxed{\text{even}}
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the composite function $f(f(g(f(g(f(x)) ))))$ is $\boxed{\text{even}}$.
</Final_answer>
</think>