TuringTest, functions?

- anonymous

TuringTest, functions?

- jamiebookeater

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- TuringTest

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?

- anonymous

Yes

- TuringTest

good
now, is x a function of y ?

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## More answers

- anonymous

No

- anonymous

Cause x is the dependent variable

- TuringTest

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

- anonymous

y=x+7
y-7=x?

- TuringTest

so is x a function of y?

- TuringTest

can we switch their places?

- anonymous

They should work both ways.

- TuringTest

right, so in
y=x+7
we can write one variable as a function of the other at will
x=y-7
so we can change the way we look at this and call it
f(x)=x+7=y
or
g(y)=y-7=x
they are both functions and both equivalent.
now trickier:
what about
y=x^2
is y a function of x ?

- anonymous

Yes?

- TuringTest

yes, each x gives exactly one y
now the hard part :P
is x a function of y?

- anonymous

No, cause each y gives us 2 x

- anonymous

It could be a function of yx

- TuringTest

how so? elaborate ;)

- anonymous

|dw:1326051567892:dw| every element of a is maped to an element of b

- anonymous

y=x^2
yx=x
Therefore each y gives us 2 x
It is a function of yx

- TuringTest

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?

- anonymous

y/x=x actually

- anonymous

Lol i rushed it

- TuringTest

that's not solved for x, you have x on both sides :/

- TuringTest

take the square root of both sides

- anonymous

y1/2=x?

- TuringTest

is the 1/2 and exponent?

- TuringTest

an exponent*

- anonymous

Yes, didnt you say that yesterday?

- TuringTest

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...

- anonymous

ALright

- TuringTest

plus or minus...

- anonymous

What?

- TuringTest

when you solve\[x^2=4\]there are two answers, remember?
one positive and one negative.
or did you not know that?

- anonymous

No, why?

- pokemon23

TURNING TEST BUDDY :D

- TuringTest

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.

- TuringTest

hi pokemon :)

- anonymous

ooooh

- anonymous

since negative multiplied by negative is positive

- TuringTest

exactly :D

- TuringTest

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...

- TuringTest

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 ;)

- TuringTest

so that said, back on topic\[r^2=t\]is t a function of r?
is r a function of t?

- anonymous

r is a function of t but not vice versa

- TuringTest

actually vice versa, but not vice versa lol

- TuringTest

which is a function of what? careful.

- anonymous

So t is a function of r?

- TuringTest

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

- anonymous

Ohh

- TuringTest

each positive value of t will give two real answers*

- anonymous

Yeah, throw me another one

- TuringTest

what about\[r=\sqrt t\]???

- anonymous

sqrt(t) cant be the dependent variable

- TuringTest

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.

- anonymous

Oh, so if the +/- sign is gone then it can me the other way around?

- TuringTest

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

- TuringTest

yes
because we are only looking at the positive root
I'll draw...

- TuringTest

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:dw|I'll draw in r=sqrt(t)...

- TuringTest

Actually first the graph of r^2=t:|dw:1326054064202:dw|

- TuringTest

Now the graph of just\[r=\sqrt t\]|dw:1326054150478:dw|see 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.

- TuringTest

|dw:1326054308550:dw|^not a function|dw:1326054323340:dw|^is a function

- anonymous

because of the plus/minus it can be
|dw:1326054372813:dw|
or
|dw:1326054351716:dw|
Depending on if its negative or positive

- TuringTest

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...

- anonymous

So that was wrong?

- TuringTest

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:dw|you would agree that y is a function of x, yes?

- anonymous

No its not

- TuringTest

because it passes the vertical line test|dw:1326054781944:dw|but what if we changed the coordinates on the graph
put x vertical and y horizontal|dw:1326054851735:dw|even 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

- anonymous

ut its x, not y. Shouldnt it be horizontal line test?

- TuringTest

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:dw|same result, we see that y is a function of x, but not the other way around.

- TuringTest

so another way to see it, is|dw:1326055195330:dw|is y a function of x?

- anonymous

Yes

- TuringTest

good
is x a function of y?

- anonymous

no

- TuringTest

show please

- anonymous

|dw:1326055271855:dw|

- TuringTest

perfect :)

- anonymous

:D

- TuringTest

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 ?

- anonymous

y is a function of x and x is a function of y

- TuringTest

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...

- TuringTest

r | t
---
0 | 1
2 | 3
5 | 9
7 | 3
4 | -14
is r(t) a function?
is t(r) a function?

- anonymous

No
Yes

- TuringTest

correct!
very nice, I think you understand the concept of functions :D

- anonymous

Seriously? :D

- TuringTest

yes, you are right :D
care to state your rationale?

- anonymous

Rationale?

- TuringTest

why? your reason.

- TuringTest

how did you know?

- anonymous

Oh, for the answer?

- TuringTest

yeah
r | t
---
0 | 1
2 | 3
5 | 9
7 | 3
4 | -14
is r(t) a function?
is t(r) a function
WHY?

- anonymous

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

- TuringTest

right, so r(t) does not represent a function, where t(r) does
perfect!!!!

- anonymous

Yeah!!! :DDD

- TuringTest

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 :)

- anonymous

Ok :D Im going to take a 30min break now :)

- TuringTest

well deserved, take your time :D

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