A) Domain and Range in Set Notation:
- Domain: [tex]\(\{ x \in \mathbb{R} \}\)[/tex]
- Range: [tex]\((-\infty, 1]\)[/tex]
B) Intervals where the graph is positive is [tex]\(\left( -\frac{7}{3}, -\frac{5}{3} \right)\)[/tex]
C) Intervals where the graph is negative is [tex]\((-\infty, -\frac{7}{3}) \cup (-\frac{5}{3}, \infty)\)[/tex]
From the question, we have the following parameters that can be used in our computation:
[tex]f(x) = -3|x + 2| + 1[/tex]
The domain of the function is all real numbers since the absolute value function is defined for all real numbers.
[tex]\[\text{Domain} = \{ x \in \mathbb{R} \}\][/tex]
To find the range, we need to understand the behavior of the function.
The function -3|x + 2| + 1 shifts the graph of -3|x| horizontally by 2 units to the left and vertically by 1 unit upwards.
The minimum value of -3|x + 2| is 0 (when x = -2).
So, the maximum value of f(x) is:
[tex]\[f(-2) = -3|(-2) + 2| + 1 = 1\][/tex]
As |x + 2| increases, -3|x + 2| becomes more negative, so the function has no lower bound.
So, the range is:
[tex]\[\text{Range} = (-\infty, 1]\][/tex]
To find where the graph is positive, solve f(x) > 0
[tex]\[-3|x + 2| + 1 > 0\][/tex]
[tex]\[-3|x + 2| > -1\][/tex]
[tex]\[|x + 2| < \frac{1}{3}\][/tex]
So,
[tex]\[-\frac{1}{3} < x + 2 < \frac{1}{3}\][/tex]
[tex]\[-2 - \frac{1}{3} < x < -2 + \frac{1}{3}\][/tex]
[tex]\[-\frac{7}{3} < x < -\frac{5}{3}\][/tex]
So, the interval where the graph is positive is:
[tex]\[\left( -\frac{7}{3}, -\frac{5}{3} \right)\][/tex]
To find where the graph is negative, solve f(x) < 0
[tex]\[-3|x + 2| + 1 < 0\][/tex]
[tex]\[-3|x + 2| < -1\][/tex]
[tex]\[|x + 2| > \frac{1}{3}\][/tex]
So,
[tex]\[x + 2 < -\frac{1}{3} \quad \text{or} \quad x + 2 > \frac{1}{3}\][/tex]
[tex]\[x < -2 - \frac{1}{3} \quad \text{or} \quad x > -2 + \frac{1}{3}\][/tex]
[tex]\[x < -\frac{7}{3} \quad \text{or} \quad x > -\frac{5}{3}\][/tex]
So, the intervals where the graph is negative are [tex]\[(-\infty, -\frac{7}{3}) \cup (-\frac{5}{3}, \infty)\][/tex]