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- anonymous

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- anonymous

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- DebbieG

*getting a sore neck* lol
OK, F(x) = 3x + 6
We've done other problems like this, so what's the first step? Put the question in y=f(x) notation, e.g.
y = 3x + 6

- anonymous

move the x and y

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## More answers

- DebbieG

Now, an inverse swaps the domain and range of a function. So this is where NORMALLY we would change the x to a y and the y to an x.
However, as we've talked about before, your teacher seems to want you to give the inverse as a function of y (not standard... you might ask about that???) so you should really just SOLVE FOR x.

- anonymous

remember last time you help well i got all of them wrongg -_-

- DebbieG

Well, if you mean switch the x to y and the y to x, that's hard to answer here because of your teacher's unusual notation.
Normally the inverse of the function F(x) is given by \(F^{-1}(x)\), not as \(F^{-1}(y)\). But since your teacher used the notation \(F^{-1}(y)\), I'm assuming that (s)he wants it to be a function of y.

- DebbieG

OK, 2 questions:
1. how did you do it last time? did you give your answer as a function of y, or of x?
2. Did you ask the teacher why they were all wrong? Did you get any feedback?
Because that goofy notation is really throwing me. I can explain it either way, but I'm not sure what your teacher wants.

- anonymous

did u see the pic what it telln me to solve for

- DebbieG

Yes, I saw the picture. The picture says to find "the inverse of F(x)"... that's cool, I know how to do that. But that is called \(F^{-1}(x)\), NOT \(F^{-1}(y)\) as your teacher's notation says.
\(F^{-1}(x)\) is a function of x, and that is how the inverse of a function of x is normally given.
\(F^{-1}(y)\) would be a function of y.

- anonymous

ok im lost

- DebbieG

ACTUALLY, \(F^{-1}(y)\) SHOULD be the function I get, if I take an inverse of a function of y.
E.g., if I have a function \(F(y)=3y-7\) and I take it's inverse, THEN I should get a function that I would call \(F^{-1}(y)\). The inverse of a function of x should be a function of x; the inverse of a function of y should be a function of y.
The inverse of a function of x should not be a function of y.

- DebbieG

*sigh* I know you are, and I'm sorry. Finding the inverse of this function is NOT hard. It's just the notation being used by your teacher that is making it complicated. So let me ask again:
OK, 2 questions:
1. how did you do it last time? did you give your answer as a function of y, or of x?
2. Did you ask the teacher why they were all wrong? Did you get any feedback?
Because that goofy notation is really throwing me. I can explain it either way, but I'm not sure what your teacher wants.

- DebbieG

@amistre64 @AkashdeepDeb have either of you seen this weird kind of notation for the inverse before? Any thoughts??

- amistre64

yes, there is still debate about it :)

- amistre64

the inverse of f(x) is such that the values that relate to x are a function composed of y values

- DebbieG

LOL so do you think he should just solve the equation as-is for x, and let that be the inverse, x as a function of y?

- AkashdeepDeb

How much is? F^-1(x) guys?

- amistre64

f(x) = 3x+6
f(x) is some curve in the xy plane such that for any given value of x gives us a value such that y = f(x)
y = 3x+6 , the inverse of this setup is such that: (y-6)/3 produces the values of x that relate to inputs of y

- amistre64

i believe that he should simply solve for the inverse and keep it as a function of y ...

- AkashdeepDeb

Can you please tell me the value of
F^-1(x) ?

- anonymous

can someone teach me how to solve please

- DebbieG

Thanks for the input. @AkashdeepDeb I didn't understand your question... lol.
I'll have to read up on this, as I've always taught it with switch and solve.

- amistre64

spose \(f:X\to Y\) and \(g:Y\to X\) such that \(f\) and \(g\) have an inverse relationship to each other .... etc

- amistre64

http://www.mathsisfun.com/sets/function-inverse.html

- DebbieG

@AkashdeepDeb \(f^{-1}(x)\) maps the element x in the range of F back to x in the domain of F....?? I'm not sure what you're looking for. Maybe we should take this to another post so as not to give @romanortiz65 nightmares, lol.

- anonymous

yup xD im totaly losttt

- DebbieG

@romanortiz65 here is what you need to know:
You have y = 3x+6
Solve it for x, so you will have x={some stuff that involves y}
Then that is your \(f^{-1}(y)\), eg,
\(f^{-1}(y)=\text{{some stuff that involves y}}\)

- amistre64

|dw:1377539248495:dw|

- DebbieG

Yeah, I get the picture @amistre64 .... but then you don't get that cool reflection of the function and inverse around y=x! What's the fun in that??? lol

- amistre64

whereas ...|dw:1377539344812:dw|

- amistre64

it just nitpicky notation that is all;
the inverse of a function just back tracks the solutions is all

- anonymous

people i need help xD

- DebbieG

@romanortiz65 read my post just above @amistre64 first picture (about 6 posts up from here).

- DebbieG

here is what you need to know:
You have y = 3x+6
Solve it for x, so you will have x={some stuff that involves y}
Then that is your\( f^{âˆ’1}(y)\), eg, \(f^{âˆ’1}(y)\)={some stuff that involves y}

- anonymous

x+3/6

- DebbieG

Show your steps.... I'm not sure how you got that. And it doesn't involve y. And it isn't an equation.
Just start from y = 3x+6 and solve it for x. Isolate the x on one side.
You should end up with x={some stuff that involves y}

- anonymous

6-y=3x

- anonymous

is it kinda like this ?

- anonymous

hey @DebbieG

- DebbieG

You have y = 3x+6
You want x by itself. So first, SUBTRACT 6 from both sides. What do you get?

- anonymous

-6y=3x

- DebbieG

Well, what you just did was SUBTRACT 6 from the RHS, but you MULTIPLIED BY -6 on the LHS. Try again. :)
You have y = 3x+6
SUBTRACT 6 from both sides.... what do you get?

- DebbieG

(just in case it isn't obvious, RHS means right hand side, LHS means left hand side)

- anonymous

-6y=x-3

- anonymous

hey @DebbieG

- anonymous

@Luigi0210 @mathstudent55

- anonymous

can you see the pic all the way to the top

- DebbieG

I'm sorry, @romanortiz65 .... for some reason I'm not getting the notifications to this thread immediately.
Yes, I can see the pic... the equation is y = 3x+6 , right??
From: y = 3x+6 when you subtracted 6 from both sides, HOW are you getting -6y=x-3??

- anonymous

ohh snaps xD y=3x

- DebbieG

If you have a, and I say "subtract 6 from a", you get a-6
Now: you have y = 3x+6
Subtract 6 from 3x + 6: ----> 3x + 6 -6 = 3x
Subtract 6 from y: -----> ??

- DebbieG

huh? what is?? is the function we are starting with y = 3x+6 ????

- anonymous

y=3x+6
-6 -6
y=3x

- DebbieG

@romanortiz65 ....... what is y minus 6??
How is y minus 6 =y??

- anonymous

-6y=3x

- DebbieG

-6y is "y TIMES (-6)". It is NOT y-6
If you have 8 apples, and I take 6 away, do you have -48 apples?

- anonymous

2 apples

- DebbieG

Yes, because "a minus 6" is "a - 6".... it is not "-6a" right??

- anonymous

ohhhh -6=3x

- DebbieG

Now: you have y = 3x+6
Subtract 6 from 3x + 6: ----> 3x + 6 -6 = 3x
Subtract 6 from y: -----> ??

- DebbieG

where
did
your
y
go???

- DebbieG

NO. stop. We aren't to that point yet. (and that's wrong, anyway. and it isn't a function of y!!)
you have y = 3x+6
Subtract 6 from 3x + 6: ----> 3x + 6 -6 = 3x
Subtract 6 from y: -----> ??
WHAT do you get when you subtract 6 from y?

- anonymous

y-6

- DebbieG

YES!! YES YES YES! :) ok so now you have, what, for your equation?

- anonymous

y-6=3x

- DebbieG

PERFECT. now we are almost there. You want the x all alone. Now what do you do, to get that coefficient of 3 to go away?

- anonymous

i mean 3

- anonymous

-3 both sides

- DebbieG

OK, that's not as bad as when I saw "-6" lol... What do you mean, do you mean subtract 3?? That won't do it.

- anonymous

it wont ?????

- DebbieG

That will just give me, in the RHS:
3x - 3
Which is not x all alone.

- DebbieG

Hint:

- anonymous

divide 3

- DebbieG

\[10\cdot2=20\]
\[\dfrac{10\cdot2}{2}=\dfrac{20}{2}\]
\[10=10\]

- DebbieG

YES, divide by 3 on both sides!! Good job! You see why it HAS to be division, right?

- DebbieG

You are "undoing" the product of 3 and x.... and division "undoes" multiplication.

- anonymous

y-2=x?

- DebbieG

not quite. You have to divide the WHOLE LHS by the 3. Not just the 6!

- anonymous

2y-2=x

- DebbieG

hmm. No, y divided by 3 is definitely not = 2y

- anonymous

lol 3y-2=x

- DebbieG

Just LEAVE the expression on the left as-is, and put it ALL over a denominator of 3.

- DebbieG

No, because y divided by 3 is also not = 3y

- anonymous

y-2/3=x

- DebbieG

if \(t+5=7u\) then \(\dfrac{ t + 5 }{7 }=u\)

- anonymous

thats what i did

- DebbieG

No.... you have y-6=3x
if you apply my example to that, you do NOT get
\[y-\dfrac{ 2 }{3 }=x\]

- anonymous

\[f ^{-1}(y)=\frac{ y-2 }{ 3 }\]

- DebbieG

SO CLOSE. But how did your y - 6 become y - 2?

- anonymous

snapps

- DebbieG

(And, if that's ^^^ what you meant, you should have typed (y-2)/3, not y-2/3, they have different meaning :)

- anonymous

final answer is (y-2)/3?

- anonymous

@DebbieG

- DebbieG

No. If you have y - 6, and you divide that WHOLE DARN THING, the whole (y - 6) by 3, what do you get??
You don't get (y-2)/3......

- anonymous

my bad ;( damn

- DebbieG

's ok, scroll up and look at my example again, with t and 5 and u.

- anonymous

(y-6)/3 like that

- DebbieG

YES!! yes yes and again yes. that is your f^(-1)(y).

- anonymous

YEAHHHHHHHHH xD

- DebbieG

\(x=\dfrac{ y - 6 }{3 }\)

- anonymous

wtf its not likethis \[\frac{ (y-6) }{ 3 }\]

- DebbieG

THAT'S FINE TOO... they are equivalent.
When you have it as a fraction, "up and down" so to speak, the "( )" aren't critical. The placement of the terms in the numerator and the other terms in the denominator is CLEAR.
It's only when you do a fraction in "text", like
y - 6/3 vs. (y-6)/3
that the ( ) are critical, because otherwise the meaning isn't clear.

- DebbieG

Sorry, didn't mean to throw you for a loop!!

- anonymous

ok so which one is right

- DebbieG

\[\Large \dfrac{y - 6 }{3 }=\dfrac{ (y - 6) }{3 }\]
BUT
\[\Large y-6/3=y-\frac{ 6 }{3 }\neq \frac{ y-6 }{ 3 }\]

- DebbieG

Like I said, they are equivalent.
\[\Large \dfrac{y - 6 }{3 }=\dfrac{ (y - 6) }{3 }\]
But it never HURTS to use the ( ) so I would use them. Definitely if you are entering as text.

- anonymous

oh okay thanks

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