# Onto Function

A function f: A -> B is called an onto function if the range of f is B. In other words, if each b ∈ B there exists at least one a ∈ A such that.
f(a) = b, then f is an on-to function. An onto function is also called surjective function.
Let A = {a1, a2, a3} and B = {b1, b 2 } then f : A -> B. ## Examples on onto function

Example 1: Let A = {1, 2, 3}, B = {4, 5} and let f = {(1, 4), (2, 5), (3, 5)}. Show that f is an surjective function from A into B.
Solution: Domain = {1, 2, 3} = A
Range = {4, 5}
The element from A, 2 and 3 has same range 5.
So f : A -> B is an onto function.

Example 2: State whether the given function is on-to or not. f : R -> R defined by f(x) = 1 + x2.
Solution: f(x) = 1 + x 2
Let x = 1
f(1) = 1 + 1 2
f(1) = 1 + 1
f(1) = 2 ----(equation 1)
Now, let x = -1
f(-1) = 1+ (-1) 2
= 1 + 1
f(-1) = 2 -----(equation 2)
but 1 ≠ -1
∴ f is not one to one function.
If f is onto then O ∈ R
f(x) = 0
there4; 1 + x2 = 0
there4; x2 = -1
x = ± √-1
but x ∈ R
∴ f is not on-to function

Example 3: Determine which of the following functions f : R -> R are onto i. f(x) = x + 1
Solution: f(x) = x + 1
x = 1 ; f(x) = 1 + 1
f(1) = 2
x = -1 ; f(-1) = -1 + 1
f(-1) = 0
x = 0; f(x) = x + 1
f(0) = 0 + 1
f(0) = 1
From the above we can see that domain = {-1, 0, 1} and
range = {2, 0, 1}
So, the given function is on-to.

ii) f(x) = |x| + x
Solution: f(x) = |x| + x
For x = 0 ; f(0) = 0
For x = -1; f(-1) = |-1| + 1
= 1 + 1
f(-1) = 2
for x = 1; f(1) = |1| + 1
= 1 + 1
f(1) = 2
Domain = {0, -1, 1.}
Range = {0, 2, 2. }
So, the given function is not on-to.

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