given f(x) = (2x-2)/4 solver for f^1(3)

- anonymous

given f(x) = (2x-2)/4 solver for f^1(3)

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- anonymous

:/

- anonymous

@amistre64 @phi @jim_thompson5910 @cwrw238

- amistre64

when y=3, what does x have to be?
or, for what value of x, does f(x) = 3

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## More answers

- amistre64

\[f(x)=3\]
\[f^{-1}(f(x))=f^{-1}(3)\]
\[x=f^{-1}(3)\]

- anonymous

I know that.... Solving it is my issue because I can't just plug it in

- amistre64

sure you can, i just demonstarted that f(x) = 3

- amistre64

f(x) = (2x-2)/4
3 = (2x-2)/4

- amistre64

the value of x is equal to f^-1(3)

- amistre64

even if we dont go that route, but replace x by f^-1(3)
\[f(x)=\frac{2x-2}{4}\]
\[f(f^{-1}(3))=\frac{2(f^{-1}(3))-2}{4}\]
\[3=\frac{2(f^{-1}(3))-2}{4}\]
solve for f^(-1)(3)

- anonymous

If I plug it in, I get 1.
i don't know what the ^-1 means

- amistre64

^-1 is inverse notation ....

- anonymous

I know that

- amistre64

youre giving me mixed results here ... you either know or you dont; it cant be both :/

- anonymous

Let me rephrase. I KNOW what the ^-1 IS but I don't know what to do with it

- amistre64

hmm, you recall that y = f(x) right?

- anonymous

yep, sorry for the miscommunication

- amistre64

to undo the f, we apply its inverse; so f^-1 each side
\[y=f(x)\]
\[f^{-1}(y)=f^{-1}(f(x))\]
\[f^{-1}(y)=x\]
letting y = 3
\[f^{-1}(3)=x\]
and since y = f(x), let f(x) = 3 and solve for x

- amistre64

hmm, you recall that y = f(x) right?

- anonymous

so the Y actually equals 3? Although I don't have a Y? Seems weird that they'd give me that

- amistre64

f(x) is a curve in the xy plane; for some value of x, we can measure the value of the curve, and relate it to the y axis.

- phi

you could think like this
f(x) is a rule that changes x into y
the inverse goes the other way, given y, you find x

- amistre64

|dw:1378406020372:dw|

- anonymous

Why is this question so complex? s: I have at least three others like this

- amistre64

its trying to get you to understand some basic concepts which havent really been addressed that clearly to you

- amistre64

a function has a specific value attrbuted to a given value of x, we can measure that value f(a), using the y axis such that y = f(a), for some x=a

- anonymous

Exactly. And to anyone's surprise, this actually isn't in my lesson. It explains functions and inverse but it doesn't give examples. Math is my worst subject so I don't understand any of it.

- amistre64

lets define a function as the path to get from the couch to the refridderator ... the inverse function is how you get back to the couch again

- amistre64

|dw:1378406257168:dw|

- amistre64

thats all in inverse does is get you back to where you started

- amistre64

so if we made it to y=f(x)=3. then the inverse process f^-1(3) will get us back to the value of x that got us there to start with

- anonymous

i know. It's the reverse basically

- amistre64

soo, \[3 = \frac{2x-2}{4}\]
what value of x got us to f(x) = 3?

- anonymous

the inverse?

- anonymous

what you're asking is unclear

- amistre64

solve for x ...

- anonymous

I don't know how e_e I've never had to solve for x with a fraction

- amistre64

i would suggest getting rid of the /4 by multiplying each side by 4 ... this is just your algebra rules in practice is all; the name of the variable is rather inconsequential

- anonymous

but the X is on the right side of the equal sign, rather than the left. What other "side" and I supposed to multiply besides 3? 3*4 is 12, okay... What about that remaining?
I'm sure there's a better way to explain this..

- amistre64

do you agree that a=b is the same as b=a ? does it matter what side of the equal side anything is really on? It matter when programing a computer program, but only because the computer is very precise in how it reads the program

- amistre64

\[3 = \frac{2x-2}{4}\]
\[3*4 = \frac{2x-2}{4}*4\]
\[12 = \frac{2x-2}{1}*1\]
\[12 = 2x-2\]etc...

- amistre64

what would you suggest we do next?

- anonymous

No, the side doesn't matter but my question still remains the same as to what else is there to multiply besides 3.
From there, my only guess is to add 12 to both sides the divide by 2(x)

- anonymous

edit: why is 4*4 = 1? S: you only took care of the denominator and not the numerator equation

- amistre64

4/4 = 1 this is pretty basic stuff that i am assuming you already know
\[\frac{2x-2}{4}*4\]
\[(2x-2)*\frac{1}{4}*4\]
\[(2x-2)*\frac{4}{4}\]
\[(2x-2)*\frac{1}{1}\]
\[(2x-2)*{1}\]
\[2x-2\]

- anonymous

What the hell....... You JUST said "i would suggest getting rid of the /4 by multiplying each side by 4" NOW you're saying DIVIDE? Could you /please/ make up your mind? Division and multiplication are two different things!

- amistre64

i just multiplied the right side by 4, and drew out all the little intermediary processes that show how the right side gets worked out ....

- anonymous

I'd like to get help from someone else if you don't mind.

- amistre64

i dont mind

- phi

when you have time, watch Khan's videos
start with
http://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/why-of-algebra/v/why-we-do-the-same--thing-to-both-sides--simple-equations
they are all short

- anonymous

Halfway into the video, I look at the title.
I know things should be done to both sides. @phi I'm not really sure why you sent that to me. Amistre is making the question more confusing and it's clear. First he says multiplication, then he tells me division.
I know it's against the site, but I'm really only looking for an answer. I have a lot more questions to go and I can't continue to be stuck on this one.

- anonymous

and being unclear*
I just want an answer x.x

- phi

The "answer" is knowing how to do the problem
if you don't know how, you should try to learn how.

- anonymous

._________________________.
Really. Really?
Nobody is getting anywhere with this so I'm closing it.

- phi

if you know what's in that video, move on to more complicated ones
http://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/equations_beginner/v/two-step-equations
at some point you will find where the videos are teaching you what you don't yet know.

- amistre64

the problem already had a division by 4 on the right side; to get rid of it you need to multiply both sides by 4. These are basic and fundamental algebra skills.

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