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3 years ago

State the domain and range of the relation. Determine whether the relation represents a function.

Mathematics

1 answer:

dangina [55]3 years ago

4 0

Answer:

Domain {-2,0,2}

Range {-2,0,2}

Relation is a Function

Step-by-step explanation:

We are given a relation:

{ (-2,-2) , (0,0) , (2,2) }

Domain can be defined as the all possible values of x for a relation. It is considered as a set of all first values of the ordered pairs of a given relation.

Domain of the given relation is {-2,0,2}

Range can be defined as all possible value of y which corresponds to the values of x in the domain. It is considered as a set of all second values of the ordered pairs of a given relation.

Range of the given relation is {-2,0,2}

A relation is a function if only there is one value of y for each value of x. If in the set of ordered pair of the relation, the value of x gets repeated, then the relation is not a function.

As no values of x are getting repeated, the relation is a function.

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3 years ago

Can someone please help and explain its due in a littleee

jek_recluse [69]

$f(x) = {x}^{2} - 3x + 2$

put x = (-1)

$f( - 1) = {( - 1)}^{2} - 3 \times ( - 1) + 2 \\ \\ 1 + 3 + 2 \\ \\ 6.$

put x = (z)

$f(z) = {z}^{2} - 3 \times z + 2 \\ \\ {z}^{2} - 3z + 2 \\ \\ {z}^{2} - 2z - z + 2 \\ \\ z(z - 2) - 1(z - 2) \\ \\ (z - 1)( z -2 ) \\ \\ (z - 1) = 0 \: \: , \: \: (z - 2) = 0\\ \\ z = 1 \:, \: 2$

put x = (x+1)

$f(x + 1) = {x}^{2} - 3x + 2 \\ \\ {(x + 1)}^{2} - 3 (x + 1) + 2 \\ \\ {x}^{2} + 1 + 2x - 3x - 3 + 2 \\ \\ {x}^{2} - x \\ \\ x(x - 1) \\ \\ x(x - 1) = 0 \\ \\ x - 1 = \frac{0}{x} \\ \\ x - 1 = 0 \\ \\ x = 1.$

put x = (√2 + 1 )

$f( \sqrt{2} + 1) = {x}^{2} - 3x + 2 \\ \\ {( \sqrt{2} + 1) }^{2} - 3( \sqrt{2} + 1) + 2 \\ \\ 2 + 1 + 2\sqrt{2} - 3 \sqrt{2} - 3 + 2 \\ \\ 3 - 1 \sqrt{2} - 1 \\ \\ 2 - \sqrt{2} \\ \\ 2 - 1.4 (optional \: \: \: steps)\\ \\ 0.6 \: \:$

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2 years ago