An interval that represents a set of members by including both lower and higher values is called a closed interval.
According to the Set Theory, the members (or elements) are collected to represent their collection as a set. Actually, all elements lie between two members. Hence, all members can be expressed as an interval of two members. In this case, a set should be expressed by including the lower and higher value members and it is called a closed interval.
$x \ge a$ and $x \le b$
The two inequalities tell the following two factors in mathematical form.
For our convenience, the two mathematical statements can also be written as follows.
$a \le x \le b$
This single mathematical inequality expresses that the value of $x$ lies between $a$ and $b$, and also equals to $a$ and $b$. Hence, this mathematical inequality is written as a closed interval between $a$ and $b$.
In mathematics, a closed interval is represented in two different ways.
A closed interval is represented in graphical system by considering the following two factors.
A closed interval is represented in mathematical form by considering the following two factors.
Therefore, a closed interval between $a$ and $b$ is written as $x \,∈\, \big[a, b\big]$
As per the set builder notation, a closed interval between $a$ and $b$ is written in the following forms.
$(1).\,\,\,$ $\Big\{x \,\,|\,\, x \,∈\, R \,\, and \,\, a \, \le \, x \, \le \, b\Big\}$
$(2).\,\,\,$ $\Big\{x \,:\, x \,∈\, R \,\, and \,\, a \, \le \, x \, \le \, b\Big\}$
Evaluate $f(x)$ if $f(x) = x+1$ where $x \,∈\, \big[2, 5\big]$
Let’s understand the concept of closed interval from this simple example. In this problem, the value of the function has to evaluate for every value of $x$ but the value of $x$ should be from $2$ to $5$. So, find the value of function $f(x)$ by substituting the value of $x$ from $2$ to $5$.
$f(2) \,=\, 2+1 = 3$
$f(3) \,=\, 3+1 = 4$
$f(4) \,=\, 4+1 = 5$
$f(5) \,=\, 5+1 = 6$
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