Business MathematicsNEB 2080 (old course)
a) Rewrite 2x 5 3 without absolute value sign. [3] b) Find the domain and the range of the following relation R = \(1,1), (2,4), (3,9), (4,16)\ [3]
6- a) Rewrite without absolute value sign. [3]
- b) Find the domain and the range of the following relation [3]
Answer
a) Rewriting without absolute value sign
The inequality involves an absolute value. To rewrite it without the absolute value, we use the definition of absolute value inequalities. The general form (where ) can be rewritten as .
Applying this to :
Now, we solve the compound inequality step-by-step:
Add 5 to all parts of the inequality to isolate the term with : Simplifying:
Divide all parts by 2 to solve for : Simplifying:
Thus, the inequality can be rewritten without the absolute value sign as:
b) Finding the domain and range of the relation
Domain
The domain of a relation is the set of all first elements (input values) of the ordered pairs in the relation.
Given the relation:
The first elements of the ordered pairs are: 1, 2, 3, 4.
Thus, the domain of is:
Range
The range of a relation is the set of all second elements (output values) of the ordered pairs in the relation.
The second elements of the ordered pairs are: 1, 4, 9, 16.
Thus, the range of is:
Additional Explanation (for deeper understanding)
Absolute Value Inequality
The inequality can be interpreted as the distance between and 0 on the number line is less than or equal to 3. This means lies between and , which is why we rewrite it as a compound inequality.
Relation and Domain/Range
The relation is given as a set of ordered pairs. Each ordered pair means that is mapped to . The domain is the set of all values, and the range is the set of all values.
Observing the Pattern
If we observe the relation more closely, we notice that the second elements (outputs) are perfect squares of the first elements (inputs): This suggests that the relation represents the function for . However, since the domain is explicitly given as , the range is the set of squares of these values.
Discussion
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