1. The problem asks to find the interval where the function $f(x) = 3|x + 1| - 2$ is increasing. 2. Recall that the absolute value function $|x|$ is defined as: $$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$ 3. For $f(x) = 3|x + 1| - 2$, the critical point is where the expression inside the absolute value changes sign, i.e., at $x = -1$. 4. Consider two cases: - For $x \geq -1$, $|x + 1| = x + 1$, so $$f(x) = 3(x + 1) - 2 = 3x + 3 - 2 = 3x + 1$$ - For $x < -1$, $|x + 1| = -(x + 1) = -x - 1$, so $$f(x) = 3(-x - 1) - 2 = -3x - 3 - 2 = -3x - 5$$ 5. Now, find the derivative (rate of change) in each interval: - For $x \geq -1$, $f'(x) = 3$ (positive, so $f$ is increasing here) - For $x < -1$, $f'(x) = -3$ (negative, so $f$ is decreasing here) 6. Therefore, the function is increasing on the interval $]-1, \infty[$. 7. The correct answer is A) $]-1, \infty[$.