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TuringTest Group TitleBest ResponseYou've already chosen the best response.3
Right, so now that we've covered a lot of that let's look for something tricky: starting simple: y=x+7 is y a function of x?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
good now, is x a function of y ?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Cause x is the dependent variable
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
x is the independent variable, but the independent variable can be a function of the dependent variable sometimes. It can work both ways. Not always, though. You must check. to find out try solving for x above
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
y=x+7 y7=x?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
so is x a function of y?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
can we switch their places?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
They should work both ways.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
right, so in y=x+7 we can write one variable as a function of the other at will x=y7 so we can change the way we look at this and call it f(x)=x+7=y or g(y)=y7=x they are both functions and both equivalent. now trickier: what about y=x^2 is y a function of x ?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yes, each x gives exactly one y now the hard part :P is x a function of y?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
No, cause each y gives us 2 x
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
It could be a function of yx
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
how so? elaborate ;)
 2 years ago

zbay Group TitleBest ResponseYou've already chosen the best response.0
dw:1326051567892:dw every element of a is maped to an element of b
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
y=x^2 yx=x Therefore each y gives us 2 x It is a function of yx
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
nice try but not, the algebra is wrong btw zbay's drawing show essentially the same concept as the tables I made y=x^2 solve for x do you know how?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
y/x=x actually
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Lol i rushed it
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
that's not solved for x, you have x on both sides :/
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
take the square root of both sides
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
is the 1/2 and exponent?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
an exponent*
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Yes, didnt you say that yesterday?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yes, just making sure, you should write it as y^(1/2)=x to avoid confusion that is right but you forgot one little detail about taking a square root of both sides of an equation...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
plus or minus...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
when you solve\[x^2=4\]there are two answers, remember? one positive and one negative. or did you not know that?
 2 years ago

pokemon23 Group TitleBest ResponseYou've already chosen the best response.1
TURNING TEST BUDDY :D
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
look at two different ways of getting the answer:\[x^2=y\]look at y=4:\[2^2=4\]\[(2)^2=4\]so x=2 and x=2 BOTH correspond to y=4 This is fine for y as a function of x, but means that x cannot be a function of y, because one y corresponds to multiple x.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
hi pokemon :)
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
since negative multiplied by negative is positive
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
exactly :D
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
so whenever you have to take the square root like that we must write\[x^2=y\to x=\pm\sqrt y\]which shows that x is not a function of y sometimes we can ignore this and only look at the positive root to make it a function, but we may have to adjust things accordingly to do that. Here is something many tutors on Open Study still don't know so listen up...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
If you are asked What is \[\sqrt4\]there is only one answer, 2 but if you are are asked \[x^2=4\]what is x? you must look at both plus and minus:\[x^2=4\to x=\pm2\] Many tutors here think that you always need +/ when you look at a square root. The truth is you only do that when you have to TAKE the square root of both sides of an equation. There, be a step ahead ;)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
so that said, back on topic\[r^2=t\]is t a function of r? is r a function of t?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
r is a function of t but not vice versa
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
actually vice versa, but not vice versa lol
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
which is a function of what? careful.
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
So t is a function of r?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yes, because it's okay for t to be the dependent variable. one r gives one t If we turn it around and solve for r however, we have to use +/sqrt so that means r won't be a function of t.\[r^2=t\to r=\pm\sqrt t\]so each value of t except zero will give us two real answers. not a function
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
each positive value of t will give two real answers*
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Yeah, throw me another one
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
what about\[r=\sqrt t\]???
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
sqrt(t) cant be the dependent variable
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
actually it can because I took away the +/ sign remember what I said about how many tutors are wrong about square roots? in the case of\[\sqrt4=2\]we have one answer, but for\[x^2=4\to x=\pm2\]we have two answers. so this is equivalent to the first case, we are only looking at the /positive/ square root.
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Oh, so if the +/ sign is gone then it can me the other way around?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
it is the act of TAKING the square root that introduces the +/, which is what makes it not a function. If we don't have to actually take the square root, then it is just positive. so\[r=\sqrt t\]is a function
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yes because we are only looking at the positive root I'll draw...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
I made our graph with r as the vertical direction so we could use the vertical line test. Think of how that works.dw:1326053937990:dwI'll draw in r=sqrt(t)...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
Actually first the graph of r^2=t:dw:1326054064202:dw
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
Now the graph of just\[r=\sqrt t\]dw:1326054150478:dwsee how we just use the positive part? notice how r^2=t does not pass the vertical line test with t as the independent variable, so r is not a function of t however in our last graph we clearly do have r=sqrt(t) with r as a function ot t.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
dw:1326054308550:dw^not a functiondw:1326054323340:dw^is a function
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
because of the plus/minus it can be dw:1326054372813:dw or dw:1326054351716:dw Depending on if its negative or positive
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
but that would be splitting the graph up from it's original form if we start Let me try to convey the idea graphically and with the table...
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
So that was wrong?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
Nothong really I just want to clear up the idea of what it means to break up the function like that. say we have\[y=x^2\]dw:1326054620838:dwyou would agree that y is a function of x, yes?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
because it passes the vertical line testdw:1326054781944:dwbut what if we changed the coordinates on the graph put x vertical and y horizontaldw:1326054851735:dweven though the function hasn't changed, by changing the axis and using the vertical line test we show that y is a function of x, but not vice versa
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
ut its x, not y. Shouldnt it be horizontal line test?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
no, vertical line test for a graph of independent variable on the horizontal dependent on the vertical however doing a horizontal line test would amount to the same thing as changing the axes if you noticed ;)dw:1326055073462:dwsame result, we see that y is a function of x, but not the other way around.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
so another way to see it, isdw:1326055195330:dwis y a function of x?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
good is x a function of y?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
show please
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
dw:1326055271855:dw
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
perfect :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
last thing is easier, tables... x  y  2  2 5  1 0  3 4  6 is y a function of x ? is x a function of y ?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
y is a function of x and x is a function of y
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
excellent actually I should be more careful with my language. With the table we can only say that this is potentially a function because we don't necessarily have all the values. That's a technicality though. ok last one...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
r  t  0  1 2  3 5  9 7  3 4  14 is r(t) a function? is t(r) a function?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
correct! very nice, I think you understand the concept of functions :D
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Seriously? :D
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yes, you are right :D care to state your rationale?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
why? your reason.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
how did you know?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Oh, for the answer?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yeah r  t  0  1 2  3 5  9 7  3 4  14 is r(t) a function? is t(r) a function WHY?
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
r  t  0  1 2  3<\ 5  9 These make t have several values of the same variable (2,7) I dont see 7  3</ duplicates on the other side. 4  14
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
right, so r(t) does not represent a function, where t(r) does perfect!!!!
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Yeah!!! :DDD
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
so now you have three interpretations of a function: graphical, numerical, and mathematical. try to harmonize them in your mind, they will be your best friends :)
 2 years ago

Inopeki Group TitleBest ResponseYou've already chosen the best response.0
Ok :D Im going to take a 30min break now :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
well deserved, take your time :D
 2 years ago
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