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To determine whether a function is even, odd, or neither, we use the following definitions:

- A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \) in the domain.
- A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \) in the domain.
- If neither condition holds, the function is neither even nor odd.

We’ll evaluate each function one by one.

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1. \( f(x) = \frac{5}{2 - x^2} \)

Compute \( f(-x) \):

\[
f(-x) = \frac{5}{2 - (-x)^2} = \frac{5}{2 - x^2} = f(x)
\]

✔ Since \( f(-x) = f(x) \), this function is even.

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2. \( g(x) = \frac{5x}{2 - x^2} \)

Compute \( g(-x) \):

\[
g(-x) = \frac{5(-x)}{2 - (-x)^2} = \frac{-5x}{2 - x^2} = -\frac{5x}{2 - x^2} = -g(x)
\]

✔ Since \( g(-x) = -g(x) \), this function is odd.

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3. \( h(x) = \frac{5x}{2 - x^3} \)

Compute \( h(-x) \):

\[
h(-x) = \frac{5(-x)}{2 - (-x)^3} = \frac{-5x}{2 - (-x^3)} = \frac{-5x}{2 + x^3}
\]

Now compare to \( -h(x) \):

\[
-h(x) = -\left( \frac{5x}{2 - x^3} \right) = \frac{-5x}{2 - x^3}
\]

Is \( \frac{-5x}{2 + x^3} \) equal to \( \frac{-5x}{2 - x^3} \)? Only if \( 2 + x^3 = 2 - x^3 \), which implies \( x^3 = 0 \), i.e., only at \( x=0 \). Not true for all \( x \).

Also, is \( h(-x) = h(x) \)? Clearly not — numerator changes sign, denominator changes too.

✘ So, neither even nor odd.

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4. \( i(x) = \frac{5x}{2x - x^3} \)

First, simplify the function (if possible):

Factor denominator: \( 2x - x^3 = x(2 - x^2) \)

So,

\[
i(x) = \frac{5x}{x(2 - x^2)} = \frac{5}{2 - x^2}, \quad \text{for } x \ne 0
\]

Note: The original function is undefined at \( x=0 \), and after simplification, we must remember the domain excludes \( x=0 \).

Now compute \( i(-x) \):

\[
i(-x) = \frac{5(-x)}{2(-x) - (-x)^3} = \frac{-5x}{-2x - (-x^3)} = \frac{-5x}{-2x + x^3} = \frac{-5x}{x^3 - 2x}
\]

Factor denominator: \( x^3 - 2x = x(x^2 - 2) = -x(2 - x^2) \)

So,

\[
i(-x) = \frac{-5x}{-x(2 - x^2)} = \frac{5}{2 - x^2} = i(x) \quad \text{(since } x \ne 0\text{)}
\]

✔ So, \( i(-x) = i(x) \) → even.

> Note: Even though the simplified form looks like \( f(x) \) from problem 1, the domain of \( i(x) \) excludes \( x=0 \), while \( f(x) \) is defined at \( x=0 \). But since evenness/oddness is checked over the domain where the function is defined, and for all \( x \ne 0 \) in the domain, \( i(-x) = i(x) \), it’s still even.

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5. \( j(x) = x^2 - \frac{x}{100} - 1 \)

Compute \( j(-x) \):

\[
j(-x) = (-x)^2 - \frac{(-x)}{100} - 1 = x^2 + \frac{x}{100} - 1
\]

Compare to \( j(x) = x^2 - \frac{x}{100} - 1 \)

→ Not equal ⇒ not even.

Compare to \( -j(x) = -x^2 + \frac{x}{100} + 1 \)

→ Not equal ⇒ not odd.

✘ So, neither even nor odd.

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## ✔ Final Answers:

1. \( f(x) = \frac{5}{2 - x^2} \) → Even

2. \( g(x) = \frac{5x}{2 - x^2} \) → Odd

3. \( h(x) = \frac{5x}{2 - x^3} \) → Neither

4. \( i(x) = \frac{5x}{2x - x^3} \) → Even

5. \( j(x) = x^2 - \frac{x}{100} - 1 \) → Neither

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Let me know if you’d like to see graphs or further explanations!