Find the domain and the range of the real function f defined by \(f(x)=\sqrt {(x-1)}\)
Solution and Explanation
The given real function is .\(f(x)=\sqrt{x-1}\)
It can be seen that \(\sqrt{x-1}\) is defined for (x-1) ≥0.
i.e., \(f(x)=\sqrt{x-1}\) is defined for x ≥1.
Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f= [1,∞).
As x ≥1
⇒(x-1) ≥ 0
⇒\(\sqrt{x-1}\) ≥ 0
Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f= [0,∞).
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Concepts Used:
Relations and functions
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Representation of Relation and Function
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
- Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}
- Roster form - {(1, 1), (2, 4), (3, 9)}
- Arrow Representation
