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Let R={0, 1, 2, 3} be the range of h(x) = x - 7. The domain of h inverse is
{-7, -6, -5, -4}
{7, 8, 9, 10}
{0, 1, 2, 3}

- anonymous

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- jamiebookeater

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- anonymous

b part

- eujc21

plug and chugg

- inkyvoyd

I don't know what either of those answers are, but I'll help if you work out the problem with me.

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## More answers

- anonymous

Help me then

- inkyvoyd

Do you remember what the domain and range of a function are?

- anonymous

Yep

- inkyvoyd

Do you know what an inverse function is?

- anonymous

Yes I do

- inkyvoyd

remember a function (here is a new abstract definition) is a operation that maps the domain onto the range. In this case you do not know what the domain of your function is.
You do know, however, that the inverse of a function maps the range of the original function on to the domain.
What answer are you leaning towards?

- anonymous

|dw:1378417045569:dw|

- inkyvoyd

alright.
You worked out the inverse of the function but it's not doing anything at all.

- inkyvoyd

Let R={0, 1, 2, 3} be the range of h(x) = x - 7.
Let's define our own domain for this function. Call it {a,b,c,d}. tell me what the inverse of a, b, c, and d are.

- anonymous

I'm confused, do I plug them into the problem as x?

- inkyvoyd

none of what I just said had anything to do with how to solve the problem except the first statement

- inkyvoyd

let me start over. The domain of h inverse is always the range of h itself. Does this make sense to you?

- anonymous

yes

- inkyvoyd

any function is a set of operations (or single operation) that takes an input and puts out an output.
domain (aka input) -> range (aka output)
an inverse function takes the output and returns the original input
original range -> original domain

- inkyvoyd

what is the original range of your function?

- anonymous

The output.. which is x-7?

- inkyvoyd

yes, but I was looking for R={0, 1, 2, 3}

- anonymous

WOW ink is trying really hard. KUDOS!

- inkyvoyd

if the range of your function is limited to 4 numbers, those 4 numbers are also the domain of the inverse function. From here on out I will refer to j(x) as inverse h(x)

- inkyvoyd

stop trying to do all that algebra you were taught in class and think for a moment about why the range of your original function is the domain of the inverse function. When you really understand why, go back to your problem and it'll make more sense

- anonymous

It's because it is the inverse of the problem which is the opposite of it

- inkyvoyd

Let R={0, 1, 2, 3} be the range of h(x) = x - 7. The domain of h inverse is?

- inkyvoyd

now, understand you could've done it the second dude's way and plugged and chugged. I have a philosophy about life and math in general though: Why give yourself more work when you gain the same amount from it? Thus, I went through all the work and made you do even more work than plugging and chugging to understand the problem. It's all cause I'm an excellent troll.

- anonymous

That is very wise, yet mean of you to make me do more work.

- inkyvoyd

#yolo

- anonymous

I was hoping you were going to make it easier for me

- inkyvoyd

I did, I explained the meaning (or made you understand) of domain and range. when you get into harder math classes, plug and chug won't work unless you know what you are doing. if you just said yes along to make me give you the answer, well, I did make it harder for you :)

- inkyvoyd

wait, tell me what answer you got

- anonymous

I got {-7, -6, -5, -4}

- inkyvoyd

nope

- anonymous

-.-

- inkyvoyd

dude

- anonymous

What, dude?

- inkyvoyd

is this written or online?

- anonymous

written and online

- inkyvoyd

the answer is 0,1,2,3

- inkyvoyd

if you want to know why I can try to explain one last time.

- anonymous

please do :)

- inkyvoyd

R={0, 1, 2, 3} be the range of h(x). Because any inverse function is a one to one relationship
That means that h(x) is also D={a,b,c,d} -> R={0,1,2,3}
therefore inverse h is {0,1,2,3}->{a,b,c,d}
the domain of inverse h is {0,1,2,3}

- anonymous

Oh I see. I was doing a whole different kind of math -_-

- inkyvoyd

do you get my logic.

- anonymous

I get your amazing logic

- inkyvoyd

Seriously? If you don't get it I'll try again. I'm not even joking, it's important that you understand this.

- anonymous

I do understand this. I'm serious. Thanks for caring though.

- inkyvoyd

alright, have fun doing the rest of the problems. :)

- anonymous

Thank you, sir :)

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