Which function has an inverse that is also a function? a.{(–1, –2), (0, 4), (1, 3), (5, 14), (7, 4)} b.{(–1, 2), (0, 4), (1, 5), (5, 4), (7, 2)} c.{(–1, 3), (0, 4), (1, 14), (5, 6), (7, 2)} d.{(–1, 4), (0, 4), (1, 2), (5, 3), (7, 1)}

Function has an inverse that is also a function : c.{(–1, 3), (0, 4), (1, 14), (5, 6), (7, 2)}

Set A to set B is said to be a function if each member of set A pairs exactly one member of set B

So, one value of x is only assigned to one value of y

A function can be expressed in the form of a cartesian diagram, sequential pairs, or arrow diagrams

If a function (f) pairs members of set A to set B, then the inverse of function f (f⁻¹) pairs members of set B to A, or easily f⁻¹ is the opposite of f

or in the form of equations

f (x) = y if and only if g (y) = x

g (y) is called the inverse of f (x)

We see the options available

From this function we can set A and B

A = {- 1,0,1,5,7}

B = {-2,4,3,14}

The inverse function becomes

{(-2, -1), (4.0), (3,1), (14,5) and (4,7)}

There is one member B that is paired with 2 members A, namely (4.0) and (4.7) so that it is not a function

set A and B

A = {- 1,0,1,5,7}

B = {2,4,5,2}

The inverse function becomes

{(2, -1), (4.0), (5.1), (4.5), (2.7)}

There is one member B that is paired with 2 members A, namely (4.0) and (4.5) so that it is not a function

set A and B

A = {- 1,0,1,5,7}

B = {3,4,14,6,2}

The inverse function becomes

{(3, -1), (4.0), (14,1), (6.5), (2.7)

Each member B is paired exactly one with member A, so that it is a function

set A and B

A = {- 1,0,1,5,7}

B = {4,2,3,1}

The inverse function becomes

{(4, -1), (4.0), (2.1), (3.5), 1.7)}

There is one member B that is paired with 2 members A, namely (4, -1) and (4.0) so that it is not a function

So the right answer is C

the function has an inverse function

the inverse of the function

the inverse of a function always a function

Keywords: function, inverse function, arrow diagrams

10 down (bang1)

5 up

5down (2)

2.5 up

2.5 down (3)

1.25 up

1.25 down (4)

now add them up : )

hoped i lol