Find the domain of following functions:
`f(x)=1/(sqrt(abs(x)-x^(2)))`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{|x| - x^2}} \), we need to determine the values of \( x \) for which the function is defined. This involves ensuring that the expression inside the square root is positive, as the square root must be defined and cannot be negative.

### Step 1: Set up the inequality
We start by requiring that the expression inside the square root is greater than zero:
\[ |x| - x^2 > 0 \]

### Step 2: Consider the cases for the absolute value
The absolute value function \( |x| \) can be defined in two cases based on the sign of \( x \).

#### Case 1: \( x \geq 0 \)
In this case, \( |x| = x \). Therefore, the inequality becomes:
\[ x - x^2 > 0 \]
Rearranging gives:
\[ -x^2 + x > 0 \quad \Rightarrow \quad x(1 - x) > 0 \]
This inequality holds when \( x \) is in the intervals where the product is positive. The critical points are \( x = 0 \) and \( x = 1 \). Testing the intervals:
- For \( x < 0 \): \( x(1 - x) < 0 \)
- For \( 0 < x < 1 \): \( x(1 - x) > 0 \)
- For \( x > 1 \): \( x(1 - x) < 0 \)

Thus, in this case, the solution is:
\[ 0 < x < 1 \]

#### Case 2: \( x < 0 \)
In this case, \( |x| = -x \). Therefore, the inequality becomes:
\[ -x - x^2 > 0 \]
Rearranging gives:
\[ -x(1 + x) > 0 \]
This inequality holds when both factors are either positive or negative. The critical points are \( x = 0 \) and \( x = -1 \). Testing the intervals:
- For \( x < -1 \): \( -x > 0 \) and \( 1 + x < 0 \) (negative)
- For \( -1 < x < 0 \): \( -x > 0 \) and \( 1 + x > 0 \) (positive)

Thus, in this case, the solution is:
\[ -1 < x < 0 \]

### Step 3: Combine the results
Now we combine the results from both cases:
- From Case 1: \( (0, 1) \)
- From Case 2: \( (-1, 0) \)

The domain of \( f(x) \) is the union of these intervals:
\[ \text{Domain of } f(x) = (-1, 0) \cup (0, 1) \]

### Final Answer
The domain of the function \( f(x) = \frac{1}{\sqrt{|x| - x^2}} \) is:
\[ (-1, 0) \cup (0, 1) \]