Question

G(x)= -1/2x+5

A) use the definition of a one-to-one function, to show that G(x) defines a one-to-one function.

B) find the inverse of g(x)

Answer 1

Given
`g(x)=(-1)/(2)x+5` we can conclude the following things:

A)
`g` is one-to-one.

B)
`g`'s inverse is given by
`g^(-1)(x)=-2(x-5)`.

Explanation:

A) If it is one-to-one, then f(a)=f(b) implies only a=b.

The function is
`g(x)=(-1)/(2)x+5`.

Let's see what
`g(a)=g(b)` implies here.

`g(a)=g(b)`

`(-1)/(2)a+5=(-1)/(2)b+5`

Subtract 5 on both sides:

`(-1)/(2)a=(-1)/(2)b`

Multiply both sides by -2:

`a=b`

This is exactly what we need to show that
`g` is one-to-one.

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If you look at a visual of
`g` on a graph. You would see it passes the horizontal line test making it one-to-one. (It is is just a diagonal line after all.)

B)

`g(x)=(-1)/(2)x+5`

`y=(-1)/(2)x+5`

The inverse is the swapping of
`x` and
`y` and then we want to remake the new
`y` the subject of the equation.

That is we have to solve the following for
`y`:

`x=(-1)/(2)y+5`

Subtract 5 on both sides:

`x-5=(-1)/(2)y`

Multiply both sides by -2:

`-2(x-5)=y`

`y=-2(x-5)`

So
`g^(-1)(x)=-2(x-5)`.

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Let's verify.

So we should get that
`g^(-1)(g(x))=x \text{ and } g^(-1)(g(x))=x`.

Let's see what happens:

`g^(-1)(g(x))`

`g^(-1)((-1)/(2)x+5)`

`-2((-1)/(2)x+5-5)`

`-2((-1)/(2)x+0)`

`-2((-1)/(2)x)`

`1x`

`x` -So that looks good so far.

`g(g^(-1)(x))`

`g(-2(x-5))`

`(-1)/(2)(-2(x-5))+5`

`(x-5)+5`

`x-5+5`

`x+0`

`x` -So that looks good.

Both ways check out so we indeed found the inverse of
`g`.