$$ f(x)=\frac{2 x^{5}-4 x^{3}+4 x}{4 x^{3}-9 x} $$ Part: $$ \mathbf{0} / 2 $$ Part 1 of 2 (a) Determine if the function is even, odd, or neither. The function is odd. The function is neither odd nor even. The function is even.

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We start with the function
$$
f(x)=\frac{2x^5-4x^3+4x}{4x^3-9x}.
$$
To determine whether $$ f $$ is even, odd, or neither, we calculate $$ f(-x) $$ and compare it with $$ f(x) $$.

**Step 1. Substitute $$-x$$ into the function.**

The function becomes
$$
f(-x)=\frac{2(-x)^5-4(-x)^3+4(-x)}{4(-x)^3-9(-x)}.
$$

**Step 2. Simplify the numerator.**

Calculate each term:
$$
2(-x)^5=2(-x^5)=-2x^5,
$$
$$
-4(-x)^3=-4(-x^3)=4x^3,
$$
$$
4(-x)=-4x.
$$
Thus, the numerator is
$$
-2x^5+4x^3-4x.
$$

**Step 3. Simplify the denominator.**

Calculate each term:
$$
4(-x)^3=4(-x^3)=-4x^3,
$$
$$
-9(-x)=9x.
$$
So the denominator is
$$
-4x^3+9x.
$$

Thus, we have
$$
f(-x)=\frac{-2x^5+4x^3-4x}{-4x^3+9x}.
$$

**Step 4. Factor out $$-1$$ from the numerator and the denominator.**

The numerator factorization:
$$
-2x^5+4x^3-4x = -\left(2x^5-4x^3+4x\right).
$$

The denominator factorization:
$$
-4x^3+9x = -\left(4x^3-9x\right).
$$

So, the function becomes
$$
f(-x)=\frac{-\left(2x^5-4x^3+4x\right)}{-\left(4x^3-9x\right)}.
$$

**Step 5. Cancel out the common factor $$-1$$.**

Cancelling the $$-1$$ from the numerator and denominator, we obtain
$$
f(-x)=\frac{2x^5-4x^3+4x}{4x^3-9x}=f(x).
$$

**Conclusion:**

Since $$ f(-x)=f(x) $$ for all $$ x $$ in the domain of $$ f $$, the function $$ f $$ is even.

**Answer:**

$$
\text{The function is even.}
$$
The function is even.

Part:
Part 1 of 2
(a) Determine if the function is even, odd, or neither.
The function is odd.
The function is neither odd nor even.
The function is even.

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Part: Part 1 of 2 (a) Determine if the function is even, odd, or neither. The function is odd. The function is neither odd nor even. The function is even.