To find the domain of the function \( f(x) = \ln(2 - x - x^2) \), we need to determine the set of values for \( x \) where the argument of the logarithm, \( 2 - x - x^2 \), is positive. This is because the natural logarithm function \( \ln(y) \) is only defined when \( y > 0 \).

So, we solve the inequality:
\[
2 - x - x^2 > 0
\]

First, rewrite the quadratic expression:
\[
-x^2 - x + 2 > 0
\]
Multiply through by \(-1\) (which reverses the inequality):
\[
x^2 + x - 2 < 0
\]

Next, factor the quadratic:
\[
(x + 2)(x - 1) < 0
\]

To find where this product is negative, we determine the roots of the quadratic equation \( (x + 2)(x - 1) = 0 \). The roots are \( x = -2 \) and \( x = 1 \).

The roots divide the number line into three intervals:
1. \( x < -2 \)
2. \( -2 < x < 1 \)
3. \( x > 1 \)

We check the sign of \( (x + 2)(x - 1) \) in each interval:
- For \( x < -2 \), both \( x + 2 \) and \( x - 1 \) are negative, so their product is positive.
- For \( -2 < x < 1 \), \( x + 2 \) is positive and \( x - 1 \) is negative, so their product is negative.
- For \( x > 1 \), both \( x + 2 \) and \( x - 1 \) are positive, so their product is positive.

Therefore, the inequality \( (x + 2)(x - 1) < 0 \) holds for \( -2 < x < 1 \).

Thus, the domain of \( f(x) = \ln(2 - x - x^2) \) is:
\[
\boxed{(-2, 1)}
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. What is the domain of \( \ln(x^2 - 4x + 3) \)?
2. How do you find the domain of a square root function?
3. What is the range of the function \( f(x) = \ln(x + 3) \)?
4. How does the quadratic discriminant help determine the domain?
5. Can a logarithmic function ever have a negative argument?

**Tip:** Always check if the expression inside a logarithm is positive to determine the function's domain.

Learn how to find the domain of the function ln(2 - x - x^2) step-by-step. This problem involves logarithmic functions and quadratic equations, focusing on determining where the argument of the natural logarithm function is positive. Suitable for high school students in Grades 10-12.

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