Work in free space at the beginning of unit

x = y + 14−14 − 14

𝑥 − 14 = 𝑦 = 𝑓−1(𝑥)

x = 2y + 14−14 − 14𝑥 − 14 = 2𝑦

𝑥

2− 7 = 𝑦 = 𝑓−1(𝑥)

Now divide both sides by 2

x = −y + 8−8 − 8𝑥 − 8 = −𝑦

−𝑥 + 8 = 𝑦 = 𝑓−1(𝑥)Now divide both sides by -1

x = −2y + 8−8 − 8𝑥 − 8 = −2𝑦

𝑥

−2+ 4 = 𝑦 = 𝑓−1(𝑥)

Now divide both sides by -2

Draw a line through restrict

the domain, we will cover this

next year. So on your

homework, after #25, do

everything except the domain

restrictions.

Remember the example from last time where we assigned each player

on the team a number?

This was a function because each player had one number, not two.

We then considered the inverse: Each number was only assigned to one

player so it too was a function.

Since they are both functions then we say that it is one-to-one.

The example in your homework where each student gets a grade is a

function, however, each grade is assigned to more than one student so it

is not a function.

Since they are not both functions it is not one to one.

pg 1167

To fill in the table we simply plug the values for x into the

equation and simplify so 𝑓 −2 = 3 −2 − 6 = −6 − 6 = −12

Remember, to find the points of the inverse we just swap our x’s and our y’s

Yes. It is a one-to-one function

because both the original equation

and the inverse are functions.

pg 1167

Yes. It is a one-to-one function

because both the original equation

and the inverse are functions.

pg 1167

No. It is not a one-to-one function

because the inverse is not a

function.

pg 1167

No. It is not a one-to-one function

because the inverse is not a

function.

pg 1167

Yes. It is a one-to-one function

because both the original equation

and the inverse are functions.

pg 1167

No. It is not a one-to-one function

because the inverse is not a

function.

sometimes

always

never

never

Sketch the graphs of the following functions on pg 1171

Now put the two graphs together for part a.

x ≥ 0

y ≥ 0

x ≥ 0y ≤ 0

x ≥ 0All real numbersx ≥ 0All real numbers