Examples

Determine if Bijective (One-to-One)
(0,9)(0,9) , (8,6)(8,6)
Step 1
Since for each value of xx there is one and only one value of yy, the given relation (0,9),(8,6)(0,9),(8,6) is a function.
The relation is a function.
Step 2
Since the relation is a function and for each value of yy there is one and only one value of xx, the given relation (0,9),(8,6)(0,9),(8,6) is a one-to-one function.
The relation is a one-to-one function.
Step 3
Every point in the range is the value of yy for at least one point xx in the domain, so this is a surjective function.
Surjective function
Step 4
Since (0,9),(8,6)(0,9),(8,6) is injective (one to one) and surjective, then it is bijective function.
Bijective function
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