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.