anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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DebbieG
  • 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
  • anonymous
move the x and y

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DebbieG
  • 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
  • anonymous
remember last time you help well i got all of them wrongg -_-
DebbieG
  • 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
  • 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
  • anonymous
did u see the pic what it telln me to solve for
DebbieG
  • 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
  • anonymous
ok im lost
DebbieG
  • 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
  • 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
  • DebbieG
@amistre64 @AkashdeepDeb have either of you seen this weird kind of notation for the inverse before? Any thoughts??
amistre64
  • amistre64
yes, there is still debate about it :)
amistre64
  • amistre64
the inverse of f(x) is such that the values that relate to x are a function composed of y values
DebbieG
  • 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
  • AkashdeepDeb
How much is? F^-1(x) guys?
amistre64
  • 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
  • amistre64
i believe that he should simply solve for the inverse and keep it as a function of y ...
AkashdeepDeb
  • AkashdeepDeb
Can you please tell me the value of F^-1(x) ?
anonymous
  • anonymous
can someone teach me how to solve please
DebbieG
  • 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
  • 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
  • amistre64
http://www.mathsisfun.com/sets/function-inverse.html
DebbieG
  • 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
  • anonymous
yup xD im totaly losttt
DebbieG
  • 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
  • amistre64
|dw:1377539248495:dw|
DebbieG
  • 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
  • amistre64
whereas ...|dw:1377539344812:dw|
amistre64
  • amistre64
it just nitpicky notation that is all; the inverse of a function just back tracks the solutions is all
anonymous
  • anonymous
people i need help xD
DebbieG
  • DebbieG
@romanortiz65 read my post just above @amistre64 first picture (about 6 posts up from here).
DebbieG
  • 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\( f^{−1}(y)\), eg, \(f^{−1}(y)\)={some stuff that involves y}
anonymous
  • anonymous
x+3/6
DebbieG
  • 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
  • anonymous
6-y=3x
anonymous
  • anonymous
is it kinda like this ?
anonymous
  • anonymous
hey @DebbieG
DebbieG
  • DebbieG
You have y = 3x+6 You want x by itself. So first, SUBTRACT 6 from both sides. What do you get?
anonymous
  • anonymous
-6y=3x
DebbieG
  • 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
  • DebbieG
(just in case it isn't obvious, RHS means right hand side, LHS means left hand side)
anonymous
  • anonymous
-6y=x-3
anonymous
  • anonymous
hey @DebbieG
anonymous
  • anonymous
@Luigi0210 @mathstudent55
anonymous
  • anonymous
can you see the pic all the way to the top
DebbieG
  • 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
  • anonymous
ohh snaps xD y=3x
DebbieG
  • 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
  • DebbieG
huh? what is?? is the function we are starting with y = 3x+6 ????
anonymous
  • anonymous
y=3x+6 -6 -6 y=3x
DebbieG
  • DebbieG
@romanortiz65 ....... what is y minus 6?? How is y minus 6 =y??
anonymous
  • anonymous
-6y=3x
DebbieG
  • 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
  • anonymous
2 apples
DebbieG
  • DebbieG
Yes, because "a minus 6" is "a - 6".... it is not "-6a" right??
anonymous
  • anonymous
ohhhh -6=3x
DebbieG
  • DebbieG
Now: you have y = 3x+6 Subtract 6 from 3x + 6: ----> 3x + 6 -6 = 3x Subtract 6 from y: -----> ??
DebbieG
  • DebbieG
where did your y go???
DebbieG
  • 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
  • anonymous
y-6
DebbieG
  • DebbieG
YES!! YES YES YES! :) ok so now you have, what, for your equation?
anonymous
  • anonymous
y-6=3x
DebbieG
  • 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
  • anonymous
i mean 3
anonymous
  • anonymous
-3 both sides
DebbieG
  • 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
  • anonymous
it wont ?????
DebbieG
  • DebbieG
That will just give me, in the RHS: 3x - 3 Which is not x all alone.
DebbieG
  • DebbieG
Hint:
anonymous
  • anonymous
divide 3
DebbieG
  • DebbieG
\[10\cdot2=20\] \[\dfrac{10\cdot2}{2}=\dfrac{20}{2}\] \[10=10\]
DebbieG
  • DebbieG
YES, divide by 3 on both sides!! Good job! You see why it HAS to be division, right?
DebbieG
  • DebbieG
You are "undoing" the product of 3 and x.... and division "undoes" multiplication.
anonymous
  • anonymous
y-2=x?
DebbieG
  • DebbieG
not quite. You have to divide the WHOLE LHS by the 3. Not just the 6!
anonymous
  • anonymous
2y-2=x
DebbieG
  • DebbieG
hmm. No, y divided by 3 is definitely not = 2y
anonymous
  • anonymous
lol 3y-2=x
DebbieG
  • DebbieG
Just LEAVE the expression on the left as-is, and put it ALL over a denominator of 3.
DebbieG
  • DebbieG
No, because y divided by 3 is also not = 3y
anonymous
  • anonymous
y-2/3=x
DebbieG
  • DebbieG
if \(t+5=7u\) then \(\dfrac{ t + 5 }{7 }=u\)
anonymous
  • anonymous
thats what i did
DebbieG
  • DebbieG
No.... you have y-6=3x if you apply my example to that, you do NOT get \[y-\dfrac{ 2 }{3 }=x\]
anonymous
  • anonymous
\[f ^{-1}(y)=\frac{ y-2 }{ 3 }\]
DebbieG
  • DebbieG
SO CLOSE. But how did your y - 6 become y - 2?
anonymous
  • anonymous
snapps
DebbieG
  • DebbieG
(And, if that's ^^^ what you meant, you should have typed (y-2)/3, not y-2/3, they have different meaning :)
anonymous
  • anonymous
final answer is (y-2)/3?
anonymous
  • anonymous
@DebbieG
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
  • anonymous
my bad ;( damn
DebbieG
  • DebbieG
's ok, scroll up and look at my example again, with t and 5 and u.
anonymous
  • anonymous
(y-6)/3 like that
DebbieG
  • DebbieG
YES!! yes yes and again yes. that is your f^(-1)(y).
anonymous
  • anonymous
YEAHHHHHHHHH xD
DebbieG
  • DebbieG
\(x=\dfrac{ y - 6 }{3 }\)
anonymous
  • anonymous
wtf its not likethis \[\frac{ (y-6) }{ 3 }\]
DebbieG
  • 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
  • DebbieG
Sorry, didn't mean to throw you for a loop!!
anonymous
  • anonymous
ok so which one is right
DebbieG
  • 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
  • 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
  • anonymous
oh okay thanks

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