Question

find the domain of $f(x) = \ln(2 - x - x^2)$.
(hint: remember that the domain of $\ln(x)$ is $x > 0$.)
\bigcirc $(-\infty, -2) \cup (1, \infty)$
\bigcirc $(-2, \infty)$
\bigcirc $(1, \infty)$
\bigcirc $(-2, 1)$
\bigcirc $(-\infty, -2)$

Explanation:

Step1: Set the argument > 0

For \( f(x)=\ln(2 - x - x^2) \), the argument of the logarithm must be positive. So we set \( 2 - x - x^2>0 \).

Step2: Rearrange the inequality

Multiply both sides by -1 (and reverse the inequality sign) to get \( x^2 + x - 2<0 \).

Step3: Factor the quadratic

Factor \( x^2 + x - 2 \) as \( (x + 2)(x - 1) \). So the inequality becomes \( (x + 2)(x - 1)<0 \).

Step4: Analyze the sign of the product

A product of two factors is negative when the factors have opposite signs.

• Case 1: \( x + 2>0 \) and \( x - 1<0 \). This gives \( x>-2 \) and \( x<1 \), so \( -2
• Case 2: \( x + 2<0 \) and \( x - 1>0 \). This gives \( x<-2 \) and \( x>1 \), which has no solution.

Answer:

\((-2, 1)\)