Example Question - interval notation
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Algebraic Set Operations Problem
\[ \begin{array}{l} \text{给定集合}A=\{x\in \mathbb{Z} | -5 \leq x \leq 10\}\text{, 集合}B=\{y |y=|x-1|\text{, 且}x \in A\}\\ \text{我们需要找到}A \cap B. \end{array} \] <p>首先, 我们需要明确集合A中x的可能值, 它包括所有满足条件的整数:</p> <p>A = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}</p> <p>然后确定B中y的可能值:</p> <p>B的值取决于A中的每个x值与1的差的绝对值. 我们列出A中值得到B中的相应值:</p> <p>B = \{ |x-1| | x \in A \} = \{6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}</p> <p>接下来，我们找出既在A中也在B中的整数:</p> <p>A \cap B = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}</p> <p>根据题目提供的选项, 得出答案为:</p> <p>D: [-1, \: 0, \: 1, \: 2]\</p> <p>但显然这个选项并不符合我们找出的交集A \cap B. 我们需要检查它是否是题目的误选, 因为没有任何提供的选项与我们找出的结果相符.</p> <p>由于题目中的选项与计算结果不一致，我们无法选择正确答案，可能是题目选项的错误或图片信息不完整.</p>
Solving a Compound Inequality
<p>Given the compound inequality \( \frac{3x+1}{2} \leq 5 + 2x \) and \( 1 + 5x \geq 8x \).</p> <p>First, solve \( \frac{3x+1}{2} \leq 5 + 2x \):</p> <p>Multiply both sides by 2 to eliminate the fraction: \( 3x+1 \leq 10 + 4x \).</p> <p>Subtract \( 3x \) from both sides: \( 1 \leq 10 + x \).</p> <p>Subtract 10 from both sides: \( -9 \leq x \) or \( x \geq -9 \).</p> <p>Now, solve \( 1 + 5x \geq 8x \):</p> <p>Subtract \( 5x \) from both sides: \( 1 \geq 3x \).</p> <p>Divide both sides by 3: \( \frac{1}{3} \geq x \) or \( x \leq \frac{1}{3} \).</p> <p>Combine the solutions: \( x \geq -9 \) and \( x \leq \frac{1}{3} \).</p> <p>So the solution to the compound inequality is \( -9 \leq x \leq \frac{1}{3} \).</p> <p>Thus, the correct answer is D. \( -9 \leq x \leq \frac{1}{3} \).</p>
Union and Intersection of Intervals
El conjunto A consiste en todos los números w tal que w es menor que 3, lo que se puede escribir utilizando notación de intervalos como: \[ A = (-\infty, 3) \] Por otro lado, el conjunto B incluye todos los números w tal que w es mayor o igual que 7, lo cual también se puede escribir en notación de intervalos como: \[ B = [7, +\infty) \] Ahora, para describir la unión de A y B (A ∪ B), debemos tomar todos los elementos que están en A o en B (o en ambos). Dado que entre A y B no hay solapamiento y existe un "gap" entre 3 y 7, la unión se escribirá como dos intervalos separados: \[ A \cup B = (-\infty, 3) \cup [7, +\infty) \] En cuanto a la intersección de A y B (A ∩ B), estamos buscando los elementos que están tanto en A como en B. Sin embargo, ya que no hay números que sean simultáneamente menores que 3 y mayores o iguales a 7, la intersección de estos dos conjuntos es el conjunto vacío, que se denota como: \[ A \cap B = \emptyset \] Así que la unión se compone de dos intervalos abiertos que no se solapan y la intersección es un conjunto vacío.
Understanding Set Intersections and Unions
Para resolver esta pregunta, primero debemos entender qué significan los conjuntos C y D dados en la imagen. El conjunto C está definido como \( C = \{z | z \geq 1\} \), lo cual significa que incluye todos los números reales mayores o iguales a 1. El conjunto D está definido como \( D = \{z | z \leq 5\} \), lo cual significa que incluye todos los números reales menores o iguales a 5. Ahora, al encontrar la intersección de C y D, denotada como \( C \cap D \), estamos buscando todos los números reales que satisfacen ambas condiciones — es decir, los números que son mayores o iguales a 1 y, al mismo tiempo, menores o iguales a 5. Por lo tanto, \( C \cap D \) incluiría todos los números reales desde 1 hasta 5, incluyendo ambos extremos. Utilizando la notación de intervalos, escribimos esto como \([1, 5]\). Por otro lado, la unión de C y D, denotada como \( C \cup D \), incluiría todos los números que están en C o en D o en ambos. Como C contiene todos los números mayores o iguales a 1 y D contiene todos los números menores o iguales a 5, la unión será todos los números reales, ya que no hay ningún número real que no esté cubierto por uno de los dos conjuntos. Utilizando la notación de intervalos, escribimos esto como \((-\infty, \infty)\) o simplemente \( \mathbb{R} \), que representa el conjunto de todos los números reales. En resumen: \( C \cap D = [1, 5] \) \( C \cup D = \mathbb{R} \)
Operations on Sets: Intersection and Union
Los conjuntos están definidos como los sigue: \( C = \{ z \mid z \geq 1 \} \) \( D = \{ z \mid z \leq 5 \} \) Para encontrar la intersección \( C \cap D \), buscamos los elementos que están en ambos conjuntos. Como \( C \) consiste en todos los números reales mayores o iguales a 1, y \( D \) consiste en todos los números reales menores o iguales a 5, la intersección \( C \cap D \) será el conjunto de todos los números reales que son tanto mayores o iguales a 1 como menores o iguales a 5. Por lo tanto, la intersección usando notación de intervalos es: \( C \cap D = [1, 5] \) Ahora, para encontrar la unión \( C \cup D \), combinamos los elementos de ambos conjuntos. En este caso, \( C \) ya contiene todos los elementos de \( D \), y además se extiende hacia el infinito en la dirección positiva. Así que la unión será todos los números mayores o iguales a 1, incluyendo todos los números que son menores o iguales a 5 (que son los mismos que ya están incluidos en \( C \)). La unión usando notación de intervalos es: \( C \cup D = [1, +\infty) \)
Solving Absolute Value Inequality
The inequality in the image is written as `8|y| ≤ 24`. Here, we want to solve for the variable `y`. The term `|y|` represents the absolute value of `y`, which means it is the distance of `y` from zero on the number line and is always nonnegative. To solve the inequality for `y`, follow these steps: 1. Divide both sides of the inequality by 8 to isolate the absolute value expression: \( |y| ≤ 24 / 8 \) \( |y| ≤ 3 \) 2. The solution to the absolute value inequality \( |y| ≤ 3 \) means that `y` can be any number within the interval [-3, 3], including -3 and 3. In interval notation, the solution is written as: \( y ∈ [-3, 3] \) This interval represents all values of `y` that satisfy the original inequality.
Solving Absolute Value Inequalities
To solve the inequality |c| - 1 ≥ 11, follow these steps: 1. Start by isolating the absolute value on one side of the inequality: |c| - 1 + 1 ≥ 11 + 1 |c| ≥ 12 2. Next, understand that an absolute value being greater than or equal to a positive number means the expression inside the absolute value can be either greater than or equal to that number or less than or equal to the negative of that number. In other words, if |a| ≥ b and b > 0, then a ≥ b or a ≤ -b. So for |c| ≥ 12, we have two cases: Case 1: c ≥ 12 Case 2: c ≤ -12 The solution to the inequality |c| - 1 ≥ 11 is therefore all c that are greater than or equal to 12, combined with all c that are less than or equal to -12. This can be expressed as: c ≥ 12 or c ≤ -12 Or, in interval notation: c ∈ (-∞, -12] ∪ [12, ∞)
Identifying Function Increase and Decrease Intervals
The image displays the graph of a function, and you are asked to determine the intervals where the function is increasing or decreasing. When a function is increasing, the y-values (the values on the vertical axis) get larger as the x-values (the values on the horizontal axis) increase. Conversely, a function is decreasing when the y-values get smaller as the x-values increase. Looking at the graph: - The function is increasing from approximately x=-4 to x=-2.5, x=-1 to x=0, and x=2 to x=3. - The function is decreasing from x=-5 to x=-4, x=-2.5 to x=-1, x=0 to x=2, and x=3 to the end of the graph at x=4. To describe these intervals, you can use interval notation, where you use parentheses to denote values that are not included in the interval (open interval) and square brackets for values that are included (closed interval). From the graph, it is not clear whether the extrema (the highest and lowest points along each interval) are included because we do not have precise points. However, since extrema are usually considered part of both increasing and decreasing intervals, if the interval is described using precise points, such points are typically included. Therefore, the intervals should be described as roughly: - Increasing: (-4, -2.5), (-1, 0), and (2, 3) - Decreasing: (-5, -4), (-2.5, -1), (0, 2), and (3, 4) If these values were precisely known from the graph, closed brackets would be used at the extrema. It is important to remember that the exact values may differ slightly, and it's often best to get these values from the context of the question or the accompanying material.
Domain and Range of a Parabolic Function
The graph shows a parabolic function that opens downwards. The domain represents all the x-values that the function includes, while the range represents all the y-values that the function can attain. Domain: The graphed function has endpoints on the x-axis at x = -4 and x = 2, with solid dots at these points indicating that these values are included in the domain. Thus, the domain of the function includes all x-values from -4 to 2. The domain in interval notation is \([-4, 2]\). Range: The highest point on the parabola is at y = 5 (also with a solid dot), which is the maximum value that the function attains, and there are no values higher than this since the parabola opens downward. The lowest y-value appears to extend toward negative infinity as the function will not have a bottom limit within the given domain. The range in interval notation is \((-\infty, 5]\). So, the domain and range of the function are: Domain: \([-4, 2]\) Range: \((-\infty, 5]\)
Set Representation in Interval Notation and Number Line Graphing
The image shows a worksheet asking to write the set in interval notation and to graph it on a number line. The set given is {x | x ≤ -1/2 or x > 3/2}. For interval notation, this set is a union of two intervals: - The first part, "x ≤ -1/2", is an interval that includes all numbers to the left of and including -1/2. In interval notation, this is written as (-∞, -1/2], using a square bracket to indicate that -1/2 is included in the set. - The second part, "x > 3/2", is an interval that includes all numbers strictly greater than 3/2. This does not include the endpoint 3/2, so we use a parenthesis. In interval notation, this is written as (3/2, ∞). The complete interval notation for the set is the union of these two intervals: (-∞, -1/2] ∪ (3/2, ∞). On the number line: a) Shade all points to the left of and including -1/2. b) Shade all points strictly to the right of 3/2. We use a closed dot or a bracket on -1/2 to show that -1/2 is included, and an open dot or a parenthesis just to the right of 3/2 to show that the interval starts after 3/2 and continues indefinitely to the right.
Absolute Value Inequality and Interval Notation
The given inequality is \( |x| \leq \frac{11}{3} \), and we are asked to express this set in interval notation and graph it on a number line. First, let's solve the absolute value inequality: An absolute value inequality like \( |x| \leq \frac{11}{3} \) means that the distance between \( x \) and 0 on the number line is less than or equal to \( \frac{11}{3} \). We break this into two separate inequalities: 1. \( x \leq \frac{11}{3} \), because \( x \) can be to the right of 0 and still be within \( \frac{11}{3} \) units of 0. 2. \( x \geq -\frac{11}{3} \), because \( x \) can also be to the left of 0 and be within \( \frac{11}{3} \) units of 0. The solution to the inequality is the set of all \( x \) that satisfy both conditions, meaning \( x \) is between \( -\frac{11}{3} \) and \( \frac{11}{3} \), inclusive of both endpoints. The interval notation for this inequality is \([- \frac{11}{3}, \frac{11}{3}] \). In terms of graphing this set on a number line, we would mark and shade the interval from \( -\frac{11}{3} \) to \( \frac{11}{3} \), including the endpoints, which are marked with a closed dot to indicate that they are part of the solution set. The grouping symbols for this interval notation would be ([]) which indicates that the set includes its endpoints. The endpoints entered into the boxes as shown in the image should be: - Left endpoint: \( -\frac{11}{3} \) - Right endpoint: \( \frac{11}{3} \) For the number line provided in part (b), the shaded region would start at \( -\frac{11}{3} \) and end at \( \frac{11}{3} \). You would ensure that there is a closed dot or a solid circle at each of the endpoints \( -\frac{11}{3} \) and \( \frac{11}{3} \) to depict that these points are included in the interval.
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