If f(x) and g(x) are non-periodic functions, then h(x)=f(g(x)) is

Options

A. non-periodic
B. periodic
C. may be periodic
D. always periodic, if domain of h(x) is a proper subset of real numbers

Answer

To determine whether the function \( h(x) = f(g(x)) \) is periodic or non-periodic given that both \( f(x) \) and \( g(x) \) are non-periodic functions, we can follow these steps:

### Step 1: Understand the Definitions
- A function \( f(x) \) is periodic if there exists a positive number \( T \) (the period) such that \( f(x + T) = f(x) \) for all \( x \).
- A function is non-periodic if no such \( T \) exists.

### Step 2: Analyze the Functions
Since both \( f(x) \) and \( g(x) \) are given to be non-periodic, it means:
- For \( f(x) \), there is no \( T_f \) such that \( f(x + T_f) = f(x) \).
- For \( g(x) \), there is no \( T_g \) such that \( g(x + T_g) = g(x) \).

### Step 3: Consider the Composition \( h(x) = f(g(x)) \)
We need to check if \( h(x) \) can be periodic. For \( h(x) \) to be periodic, there must exist a \( T \) such that:
\[ h(x + T) = h(x) \]
This translates to:
\[ f(g(x + T)) = f(g(x)) \]

### Step 4: Substitute and Analyze
If \( h(x) \) is periodic, then:
\[ g(x + T) \text{ must equal } g(x) \text{ for some } T \]
However, since \( g(x) \) is non-periodic, this cannot be true. Thus:
\[ g(x + T) \neq g(x) \]

### Step 5: Conclusion
Since \( g(x + T) \neq g(x) \), it follows that:
\[ f(g(x + T)) \neq f(g(x)) \]
This means:
\[ h(x + T) \neq h(x) \]

Thus, we conclude that \( h(x) \) cannot be periodic.

### Final Answer
Therefore, \( h(x) = f(g(x)) \) is non-periodic.