Problem 37
Question
Each of the following sets is the solution of an inequality of the form \(|x-c|<\delta\). Find \(c\) and \(\delta\). $$(-3,3)$$.
Step-by-Step Solution
Verified Answer
The values are \(c = 0\) and \(\delta = 3\).
1Step 1: Find c
To find the value of \(c\), calculate the midway point between -3 and 3. This is done by averaging -3 and 3. Therefore, \(c = \frac{-3+3}{2} = 0\). So, \(c = 0\).
2Step 2: Find delta
Next, find the value of \(\delta\). Since \(\delta\) represents the distance from \(c\) to either end of the interval, you can find \(\delta\) by subtracting the lower bound (-3) from \(c\). Therefore, \(\delta = c - (-3) = 0 - (-3) = 3\). So, \(\delta = 3\).
3Step 3: Interpret the Results
The values of \(c\) and \(\delta\) tell us that for values of \(x\) within 3 units (either less or more) of 0, \(x\) is a solution of the inequality \(|x-c|<\delta\).
Key Concepts
Absolute ValueInterval NotationMidpoint
Absolute Value
The concept of absolute value is often symbolized by vertical bars around a number, like \(|x|\). Absolute value measures the distance of a number from zero on the number line without considering which direction from zero it lies. This means that the absolute value is always a positive number or zero.
In inequalities, absolute value is often used to describe a range of values that are a certain distance away from a specific point. For the inequality \(|x-c|<\delta\), it means that any number \((x)\) should lie within \(\delta\) units of \(c\) on the number line. This forms an open interval centered at \(c\), extending \(-\delta\) to \(\delta\) around it.
- For example, \(|5| = 5\) because 5 is 5 units away from zero.
- Similarly, \(|-5| = 5\) because -5 is also 5 units away from zero.
In inequalities, absolute value is often used to describe a range of values that are a certain distance away from a specific point. For the inequality \(|x-c|<\delta\), it means that any number \((x)\) should lie within \(\delta\) units of \(c\) on the number line. This forms an open interval centered at \(c\), extending \(-\delta\) to \(\delta\) around it.
Interval Notation
Interval notation is a shorthand way of describing a set of numbers between a start point and an endpoint, which can be either included or not included in the set.
For example, the interval \((-3, 3)\) is an open interval where -3 and 3 are not part of the solution set. This notation tells us it includes all numbers greater than -3 and less than 3 but not -3 or 3 themselves.
Interval notation is very handy in mathematics, especially when expressing solutions of inequalities. It provides a clear way to express ranges of values concisely and helps in visualizing how values are distributed along the number line.
- If parentheses "( )" are used, it means the endpoints are not included. This is called an open interval.
- If brackets "[ ]" are used, the endpoints are included. This is called a closed interval.
For example, the interval \((-3, 3)\) is an open interval where -3 and 3 are not part of the solution set. This notation tells us it includes all numbers greater than -3 and less than 3 but not -3 or 3 themselves.
Interval notation is very handy in mathematics, especially when expressing solutions of inequalities. It provides a clear way to express ranges of values concisely and helps in visualizing how values are distributed along the number line.
Midpoint
The midpoint is essentially the "middle" point between two numbers on a number line. To calculate the midpoint, you simply take the average of the two numbers.
Let's consider the numbers -3 and 3 from our exercise. The midpoint is calculated as follows: \(\frac{-3+3}{2} = 0\). The result shows that 0 is the midpoint.
The concept of a midpoint is useful in various mathematical contexts, such as geometry, to find the central point between two locations or in algebra to describe the center of intervals between numbers. In inequalities involving absolute value, for example, the midpoint often serves as the value \(c\), around which a range \(\delta\) is defined.
- Add both numbers together.
- Divide the sum by 2.
Let's consider the numbers -3 and 3 from our exercise. The midpoint is calculated as follows: \(\frac{-3+3}{2} = 0\). The result shows that 0 is the midpoint.
The concept of a midpoint is useful in various mathematical contexts, such as geometry, to find the central point between two locations or in algebra to describe the center of intervals between numbers. In inequalities involving absolute value, for example, the midpoint often serves as the value \(c\), around which a range \(\delta\) is defined.
Other exercises in this chapter
Problem 37
Find the number(s) \(x\) in the interval \([0.2 \pi]\) which satisfy the equation. $$\cos 2 x=0$$.
View solution Problem 37
Sketch the set on a number line. [-2,3]\(\cup[1,5]\).
View solution Problem 37
Determine the point(s) where the line intersects the circle. $$4 x+3 y=24, \quad x^{2}+y^{2}=25$$
View solution Problem 38
Give the domain and range of the function. $$f(x)=|2 x+1|$$
View solution