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]
Answer:
[tex]\textsf{A)}\quad \begin{array}{l}\textsf{Domain:}\quad \{ x \mid x \in \mathbb{R} \}\\\textsf{Range:}\quad \{ y \mid y \leq 1 \}\end{array}[/tex]
[tex]\textsf{B)}\quad \left\{ x \mid -\dfrac{7}{3} < x < -\dfrac{5}{3} \right\}[/tex]
[tex]\textsf{C)}\quad \left\{ x \mid x < -\dfrac{7}{3} \text{ or } x > -\dfrac{5}{3} \right\}[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=-3|x+2|+1[/tex]
The given function is an absolute value function in vertex form:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Vertex form of an absolute value function}}\\\\f(x)=a|x-h|+k\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$(h,k)$ is the vertex.}\\\phantom{ww}\bullet\;\textsf{$a$ is the stretch or compression factor.}\end{array}}[/tex]
An absolute value function produces a V-shaped graph, which is symmetrical about the vertical line x = h.
If ∣a∣ > 1, the graph is vertically stretched.
If 0 < ∣a∣ < 1, it is vertically compressed.
If a > 0, the V-shaped graph opens upward.
If a < 0, the V-shaped graph opens downward.
In this case:
Therefore, the vertex of the V-shaped graph is at the point (-2, 1), and axis of symmetry is x = -2. Since a < 0, the graph opens downward, making the vertex the maximum point of the function.
The domain of any absolute value function includes all real numbers because there are no restrictions on x that would make the function undefined. In set notation, the domain of this function is expressed as:
[tex]\{ x \mid x \in \mathbb{R} \}[/tex]
As the maximum point of f(x) = -3|x + 2| + 1 is (-2, 1) and the graph opens downward, the range of the function is all real numbers less than or equal to 1. In set notation, the range of this function is expressed as:
[tex]\{ y \mid y \leq 1 \}[/tex]
[tex]\dotfill[/tex]
To determine the intervals where the graph of the function f(x) = -3|x + 2| + 1 is positive, we need to find where f(x) > 0:
[tex]-3|x + 2| + 1 > 0 \\\\\\-3|x + 2| > -1\\\\\\|x+2| < \dfrac{1}{3} \\\\\\ -\dfrac{1}{3} < x+2 < \dfrac{1}{3} \\\\\\-\dfrac{1}{3}-2 < x < \dfrac{1}{3}-2 \\\\\\-\dfrac{7}{3} < x < -\dfrac{5}{3}[/tex]
Therefore, graph is positive in the interval:
[tex]\left\{ x \mid -\dfrac{7}{3} < x < -\dfrac{5}{3} \right\}[/tex]
[tex]\dotfill[/tex]
To determine the intervals where the graph of the function f(x) = -3|x + 2| + 1 is negative, we need to find where f(x) < 0:
[tex]-3|x + 2| + 1 < 0 \\\\\\-3|x + 2| < -1\\\\\\|x+2| > \dfrac{1}{3} \\\\\\ -\dfrac{1}{3} > x+2\;\; \textsf{ or }\;\;x+2 > \dfrac{1}{3} \\\\\\-\dfrac{1}{3}-2 > x \;\; \textsf{ or }\;\; x > \dfrac{1}{3}-2 \\\\\\-\dfrac{7}{3} > x \;\; \textsf{ or }\;\; x > -\dfrac{5}{3}[/tex]
Therefore, graph is negative in the intervals:
[tex]\left\{ x \mid x < -\dfrac{7}{3} \text{ or } x > -\dfrac{5}{3} \right\}[/tex]