# 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 = {a

_{1}, a

_{2}, a

_{3}} and B = {b

_{1}, 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 + x

^{2}.

**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 + x

^{2}= 0

there4; x

^{2}= -1

x = ~+mn~ √-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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