`f(x)=sqrt(x^(2)-abs(x)-2)` . Find the domain of f(x).

Answer

To find the domain of the function \( f(x) = \sqrt{x^2 - |x| - 2} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ x^2 - |x| - 2 \geq 0 \]

Since the absolute value function \( |x| \) behaves differently based on whether \( x \) is non-negative or negative, we will consider two cases.

### Case 1: \( x \geq 0 \)

In this case, \( |x| = x \). Thus, the function simplifies to:

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

Now, we need to solve the inequality:

\[ x^2 - x - 2 \geq 0 \]

We can factor the quadratic:

\[ (x - 2)(x + 1) \geq 0 \]

Next, we find the roots of the equation \( (x - 2)(x + 1) = 0 \), which gives us \( x = 2 \) and \( x = -1 \). We will analyze the sign of the expression in the intervals determined by these roots:

1. \( (-\infty, -1) \)
2. \( (-1, 2) \)
3. \( (2, \infty) \)

Testing these intervals:

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 2)(-2 + 1) = (-4)(-1) = 4 \quad (\text{positive}) \]

- For \( -1 < x < 2 \) (e.g., \( x = 0 \)):
\[ (0 - 2)(0 + 1) = (-2)(1) = -2 \quad (\text{negative}) \]

- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (3 - 2)(3 + 1) = (1)(4) = 4 \quad (\text{positive}) \]

Thus, the expression \( (x - 2)(x + 1) \geq 0 \) holds for:

\[ x \in (-\infty, -1] \cup [2, \infty) \]

However, since we are in the case where \( x \geq 0 \), we only take the part \( x \in [2, \infty) \).

### Case 2: \( x < 0 \)

In this case, \( |x| = -x \). Thus, the function simplifies to:

\[ f(x) = \sqrt{x^2 + x - 2} \]

We need to solve the inequality:

\[ x^2 + x - 2 \geq 0 \]

Factoring gives us:

\[ (x + 2)(x - 1) \geq 0 \]

Finding the roots of the equation \( (x + 2)(x - 1) = 0 \) gives us \( x = -2 \) and \( x = 1 \). We analyze the sign of the expression in the intervals:

1. \( (-\infty, -2) \)
2. \( (-2, 1) \)
3. \( (1, \infty) \)

Testing these intervals:

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3 + 2)(-3 - 1) = (-1)(-4) = 4 \quad (\text{positive}) \]

- For \( -2 < x < 1 \) (e.g., \( x = 0 \)):
\[ (0 + 2)(0 - 1) = (2)(-1) = -2 \quad (\text{negative}) \]

- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ (2 + 2)(2 - 1) = (4)(1) = 4 \quad (\text{positive}) \]

Thus, the expression \( (x + 2)(x - 1) \geq 0 \) holds for:

\[ x \in (-\infty, -2] \cup [1, \infty) \]

However, since we are in the case where \( x < 0 \), we only take the part \( x \in (-\infty, -2] \).

### Final Domain

Combining the results from both cases, the domain of \( f(x) \) is:

\[ \text{Domain of } f(x) = (-\infty, -2] \cup [2, \infty) \]