Let `f_(1) : R to R, f_(2) : [0, oo) to R, f_(3) : R to R` be three function defined as
`f_(1)(x) = {(|x|, x < 0),(e^x , x ge 0):}, f_(2)(x)=x^2, f_(3)(x) = {(f_(2)(f_1(x)),x < 0),(f_(2)(f_1(x))-1, x ge 0):}` then `f_3(x)` is:

Options

A. even function
B. odd function
C. neither even nor odd
D. periodic function

Explanation

To find the function \( f_3(x) \), we will analyze the given functions step by step.

### Step 1: Define the functions

1. **Function \( f_1(x) \)**:
\[ f_1(x) = \begin{cases} |x| & \text{if } x < 0 \\ e^x & \text{if } x \geq 0 \end{cases} \]

2. **Function \( f_2(x) \)**:
\[ f_2(x) = x^2 \]

3. **Function \( f_3(x) \)**:
\[ f_3(x) = \begin{cases} f_2(f_1(x)) & \text{if } x < 0 \\ f_2(f_1(x)) - 1 & \text{if } x \geq 0 \end{cases} \]

### Step 2: Calculate \( f_3(x) \) for \( x < 0 \)

For \( x < 0 \):
- We have \( f_1(x) = |x| = -x \) (since \( x \) is negative).
- Therefore, \( f_2(f_1(x)) = f_2(-x) = (-x)^2 = x^2 \).

Thus,
\[ f_3(x) = f_2(f_1(x)) = x^2 \quad \text{for } x < 0. \]

### Step 3: Calculate \( f_3(x) \) for \( x \geq 0 \)

For \( x \geq 0 \):
- We have \( f_1(x) = e^x \).
- Therefore, \( f_2(f_1(x)) = f_2(e^x) = (e^x)^2 = e^{2x} \).

Thus,
\[ f_3(x) = f_2(f_1(x)) - 1 = e^{2x} - 1 \quad \text{for } x \geq 0. \]

### Step 4: Combine the results

Now we can write the complete definition of \( f_3(x) \):
\[ f_3(x) = \begin{cases} x^2 & \text{if } x < 0 \\ e^{2x} - 1 & \text{if } x \geq 0 \end{cases} \]

### Final Answer

Thus, the function \( f_3(x) \) is:
\[ f_3(x) = \begin{cases} x^2 & \text{if } x < 0 \\ e^{2x} - 1 & \text{if } x \geq 0 \end{cases} \]

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