Pre-Algebra Examples

(10.5, 11.4)

$(10.5, 11.4)$ ,

(- 8, 6.4)

$(- 8, 6.4)$

Step 1

Since for each value of

x

$x$ there is one and only one value of

y

$y$ , the given relation

(10.5, 11.4), (- 8, 6.4)

$(10.5, 11.4), (- 8, 6.4)$ is a function.

The relation is a function.

Step 2

Since the relation is a function and for each value of

y

$y$ there is one and only one value of

x

$x$ , the given relation

(10.5, 11.4), (- 8, 6.4)

$(10.5, 11.4), (- 8, 6.4)$ 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

y

$y$ for at least one point

x

$x$ in the domain, so this is a surjective function.

Surjective function

Step 4

Since

(10.5, 11.4), (- 8, 6.4)

$(10.5, 11.4), (- 8, 6.4)$ is injective (one to one) and surjective, then it is bijective function.

Bijective function

(10.5, 11.4), (- 8, 6.4)

$(10.5, 11.4), (- 8, 6.4)$

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