Ungrouped Data: When a data collection is vast, a frequency distribution table is frequently used to arrange the data. A frequency distribution table provides the...
Ungrouped Data: Know Formulas, Definition, & Applications
December 11, 2024Method of intervals: Let’s say the weather forecaster just said there is going to be a snowstorm of at least \(2\) but less than \(7\) inches of snow. What are the different levels of snow that we could expect? It could range anywhere between \(2\) and \(7\), right? This is called an interval.
An interval is the range of numbers between any two given numbers that includes all of the real numbers between them. When the forecaster said there would be at least \(2\) but less than \(7\) inches of snow, he described the amount of snow in this interval. Intervals can be written using inequalities and on number lines. They are represented using the wavy curve method or the method of intervals.
The interval notation is a method to represent an interval on a real-number line. An interval comprises the numbers lying between two specific numbers. For example, the set of numbers of \(x\) satisfying \(1 \leqslant x \leqslant 2\) is an interval that contains \(1\), \(2\), and all numbers between \(1\) and \(2\).
We can use interval notation to represent inequalities. We know an interval expressed as \(2 < x < 4\) denotes a set of numbers lying between \(2\) and \(4\).
Suppose we want to express the set of real numbers \(\left\{ {x:0 < x < 4} \right\}\) using an interval. So this can be expressed as interval notation as \(\left( {0,4} \right)\).
The set of all real numbers can be expressed as \(\left( { – \infty ,\infty } \right)\).
Following sets are the important subsets of the set \(R\) of all real numbers:
Thus, \(T = \left\{ {x:x \in R\,{\text{and}}\,x \notin Q} \right\}\).
Here, \(N \subset Z \subset Q \subset R,T \subset R,\,{\text{and}}\,N \not\subset T\).
The intervals can be classified based on the numbers set. Some sets will include the endpoints specified in the notation, while some might partially or not include the endpoints. So we have three types of intervals that are shown below.
We can follow certain rules and symbols to represent the interval notation for different types of intervals. Let us write different symbols that can be used to write a particular type of interval.
The wavy curve method also called the method of intervals, is a technique to solve the inequalities of the form \(\frac{{f\left( x \right)}}{{g\left( x \right)}} > 0\left( { < 0,\, \geqslant 0,\,{\text{or}}\, \leqslant {\text{0}}} \right)\)
We can solve the inequality with the help of the following five steps
Step 1: Factorize the given polynomials.
Step 2: Now, make the coefficient of all the variables to factors positive.
Step 3: Multiply the equation on both sides of the inequality by \(-1\) to remove the minus sign. By doing this, the inequality will reverse.
Step 4: Find the roots and asymptotes of the given inequality by equating each of the factors to \(0\)
Step 5: Now, plot the points on the number line.
Let us see an example of how this wavy curve method helps to solve the given inequality.
Example: Solve \(\left( {x – 3} \right)\,\left( {x – 6} \right)\, < 0\)
Solution: Given: \(\left( {x – 3} \right)\,\left( {x – 6} \right)\, < 0\)
Step 1: Factor the polynomials, so \(\left( {x – 3} \right)\,\left( {x – 6} \right)\, < 0\)
Step 2: Make the coefficients of variables positive i.e., \(\left( {x – 3} \right)\,\left( {x – 6} \right)\, < 0\)
Step 3: Multiply/divide both sides of an inequality by \(-1\) to remove the negative sign, so we get: \(\left( {x – 3} \right)\,\left( {x – 6} \right)\, < 0\)
Step 4: Find the roots and asymptotes of the inequality by equation each factor to zero. i.e.,
\(x – 3 = 0 \Rightarrow \,x = 3\,\)
\(x – 6 = 0 \Rightarrow \,x = 6\,\)
Step 5: Plot the points on the number line. Start with the largest factor, \(6\). Initially, a curve from the positive region must intersect the point (here, \(6\)). Now, look at the power of the respective factor. If it is odd, then change the path of the curve from their respective roots. If it is even, continue in the same region.
Hence, the required solution of the given inequality is \(3 < x < 6\) or \({x \in \left( {3,6} \right)}\).
Q.1. For a person to be selected as the leader, he should be a minimum of \(40\) years old. Give an interval representing this information.
Sol:
Let the age of the leader be \(x\).
It is given that the person’s age \(x\) should be a minimum of \(40\)
This means that \(x\) should be either greater than or equal to \(40\)
Hence, this is represented by the inequality, \({x \geqslant 40}\)
So, the required interval is \([\,40,\infty )\).
Q.2. Lola needs at least \(1400\) calories a day, but the calorie intake should not exceed \(1700\) calories. Represent the possible amount of calories she could eat using interval notation.
Sol:
Let us represent the number of calories by \(x\).
The inequality representing the possible amount of calories is \(1400 \leqslant x \leqslant 1700\)
The interval notation for this inequality is \(\left[ {1400,1700} \right]\).
Q.3. Solve \(\frac{{3x – {x^2}}}{{{{\left( {x + 4} \right)}^2}}} \geqslant 0.\)
Sol:
Step 1: Factor the polynomials, so \(\frac{{x\left( {3 – x} \right)}}{{{{\left( {x + 4} \right)}^2}}} \geqslant 0.\)
Step 2: Make the coefficients of variables positive, i.e. \(\frac{{ – x\left( {x – 3} \right)}}{{{{\left( {x + 4} \right)}^2}}} \geqslant 0.\)
Step 3: Multiply/divide both sides of an inequality by \(-1\) to remove the negative sign, so \(\frac{{x\left( {x – 3} \right)}}{{{{\left( {x + 4} \right)}^2}}} \leqslant 0.\)
Step 4: Find the roots and asymptotes of the inequality by the equation for each factor to zero. i.e., \(x=0\)
\(x – 3 = 0 \Rightarrow x = 3\)
\(x + 4 = 0 \Rightarrow x = – 4\)
Step 5: Plot the points on a graph. Start with the large factor, \(3\). Initially, a curve from the positive region intersects the point (\(3\)). Now, look at the power of the respective factor, if it is odd, then we have to change the path of the curve from their respective roots. If it is even, continue in the same region. Here, the curve would change its path at \(0\) and \(3\) because their factors have odd powers. However, at \(4\) it would not change its direction since its factor has even power.
Hence, the required solution of the given inequality is \(x \in \left[ {0,3} \right]\).
Q.4. Write the following subsets of \(R\) as interval \(\left\{ {x:x \in R,\, – 4 < x \leqslant 6} \right\}\). Also find the length of the interval.
Sol:
As we know, if \(a\) and \(b\) are two real numbers such that \(a < b\), then the set \((a,b] = \left\{ {x \in R:a < x \leqslant b} \right\}\) and \([a,b) = \left\{ {x \in R:a \leqslant x < b} \right\}\) are known as semi-open or semi-closed intervals. \((a,b]\) and \([a,b)\) are also denoted by \(]a,b]\) and \([a,b[\) respectively
Therefore, \(\left\{ {x:x \in R, – 4 < x \leqslant 6} \right\} = ( – 4,6]\)
Length of the interval \( = 6 – \left( { – 4} \right) = 10\)
Q.5. Solve \(\frac{{{{\left( {x – 1} \right)}^2}\left( {x – 2} \right)}}{{\left( {4 – x} \right){{\left( {x – 3} \right)}^5}}} \geqslant 0\) by using the wavy curve method.
Sol:
Given: \(\frac{{{{\left( {x – 1} \right)}^2}\left( {x – 2} \right)}}{{\left( {4 – x} \right){{\left( {x – 3} \right)}^5}}} \geqslant 0\)
Write all the terms in a standard form, i.e. \({\left( {x \pm \alpha } \right)}\)
So, \( – \frac{{{{\left( {x – 1} \right)}^2}\left( {x – 2} \right)}}{{\left( {x – 4} \right){{\left( {x – 3} \right)}^5}}} \geqslant 0 \Rightarrow \frac{{{{\left( {x – 1} \right)}^2}\left( {x – 2} \right)}}{{\left( {x – 4} \right){{\left( {x – 3} \right)}^5}}} \leqslant 0\)
Here, critical values are: \(x = 1,2,3,4\)
And we notice that at \(x=3,4\), the inequality is undefined.
Now, plot the critical points on the number line as and mark the signs by using the wavy curve method.
Now, we need to look at the power of the factors
If the power is odd, then we have to change the path of the curve from their respective roots while if the power is even, continuing in the same region.
Hence, the required solution of the given inequality is \(x \in ( – \infty ,2]\, \cup \left( {3,4} \right)\)
An interval comprises the numbers lying between two specific numbers. For example, the set of numbers of \(x\) satisfying \(2 \leqslant x \leqslant 4\) is an interval that contains \(2,4\), and all numbers are between \(1\) and \(2\). We can use interval notation to represent inequalities. Also, it explained how to convert an inequality to an interval. The wavy curve method, also called the method of intervals, is a technique is used to solve the inequalities of the form \(\frac{{f\left( x \right)}}{{g\left( x \right)}} > 0\left( { < 0, \geqslant 0,\,{\text{or}}\, \leqslant 0} \right)\), this method uses the fact that, \(\frac{{f\left( x \right)}}{{g\left( x \right)}}\) can only change sign at its zeros and vertical asymptotes.
Q.1. How do you solve a quadratic equation using a wavy curve?
Ans: We can use the wavy curve method to find the solutions to the questions.
Q.2. What are the types of intervals?
Ans: The intervals can be classified based on the numbers set. Some sets will include the endpoints specified in the notation, while some might partially or not include the endpoints. So we have three types of intervals that are listed below;
(i) Open-interval
(ii) Closed-interval
(iii) Semi-open or semi-closed interval
Q.3. How do you exclude numbers in interval notation?
Ans: To indicate an endpoint is included, we use a square bracket i.e. \(\left[ {\,\,\, } \right]\), to exclude an endpoint, we use parentheses i.e., \(\left( {\,\,\,} \right)\)
Q.4. How do you write numbers in interval notation?
Ans: The intervals are written with rectangular brackets or parentheses, and two numbers are separated by comma. The two numbers are called the endpoints of the interval. The number on the left denotes the least element. The number on the right denotes the greatest element.
Q.5. How do you solve quadratic inequalities by wavy curve?
Ans: The wavy curve method is comprising of simple steps;
Step 1: Factorize the given polynomials.
Step 2: Now, make the coefficient of all the variables to factors positive.
Step 3: Multiply/divide the equation on both sides of the inequality by \(-1\) to remove the minus sign.
Step 4: Find the roots and asymptotes of the given inequality by equating all the factors to \(0\)
Step 5: Now, finally plot the points on the number line and start with the largest factor. Here the curve from the positive region of the number line should intersect that point. Now let’s look at the power of the factors, and If it is odd, then we have to change the path of the curve from their respective roots while if it’s even, continuing in the same region.
Learn Everything about Subsets
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