If `f(x)=x^(3) and g(x)=x^(2)+1`, which of the following is an odd functionn (are odd functions)?
I. `f(x)*g(x)`
II. `f(g(x))`
III. `g(f(x))`

Options

A. Only I
B. only II
C. only III
D. Only II and III

Explanation

To determine which of the given functions is an odd function, we will use the definition of an odd function. A function \( h(x) \) is considered odd if it satisfies the property:

\[ h(-x) = -h(x) \]

We will analyze each function one by one.

### Step 1: Analyze \( f(x) \cdot g(x) \)

Given:
- \( f(x) = x^3 \)
- \( g(x) = x^2 + 1 \)

We need to check if \( f(x) \cdot g(x) \) is odd.

1. Calculate \( f(-x) \cdot g(-x) \):
\[ f(-x) = (-x)^3 = -x^3 \]
\[ g(-x) = (-x)^2 + 1 = x^2 + 1 \]
Therefore,
\[ f(-x) \cdot g(-x) = (-x^3) \cdot (x^2 + 1) = -x^3(x^2 + 1) = - (x^3(x^2 + 1)) = -f(x) \cdot g(x) \]
This satisfies the condition for being an odd function.

### Conclusion for I:
\( f(x) \cdot g(x) \) is an odd function.

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### Step 2: Analyze \( f(g(x)) \)

Now we check \( f(g(x)) \):

1. Calculate \( f(g(x)) \):
\[ g(x) = x^2 + 1 \Rightarrow f(g(x)) = f(x^2 + 1) = (x^2 + 1)^3 \]

2. Now calculate \( f(g(-x)) \):
\[ g(-x) = (-x)^2 + 1 = x^2 + 1 \Rightarrow f(g(-x)) = f(x^2 + 1) = (x^2 + 1)^3 \]

3. We see that:
\[ f(g(-x)) = (x^2 + 1)^3 = f(g(x)) \]
This does not satisfy the condition for being odd.

### Conclusion for II:
\( f(g(x)) \) is not an odd function.

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### Step 3: Analyze \( g(f(x)) \)

Now we check \( g(f(x)) \):

1. Calculate \( g(f(x)) \):
\[ f(x) = x^3 \Rightarrow g(f(x)) = g(x^3) = (x^3)^2 + 1 = x^6 + 1 \]

2. Now calculate \( g(f(-x)) \):
\[ f(-x) = (-x)^3 = -x^3 \Rightarrow g(f(-x)) = g(-x^3) = (-x^3)^2 + 1 = x^6 + 1 \]

3. We see that:
\[ g(f(-x)) = x^6 + 1 = g(f(x)) \]
This does not satisfy the condition for being odd.

### Conclusion for III:
\( g(f(x)) \) is not an odd function.

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### Final Conclusion:

The only odd function among the given options is:

**I. \( f(x) \cdot g(x) \)**