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TuringTest, functions?

Mathematics
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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?
Yes
good now, is x a function of y ?

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Other answers:

No
Cause x is the dependent variable
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
y=x+7 y-7=x?
so is x a function of y?
can we switch their places?
They should work both ways.
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 ?
Yes?
yes, each x gives exactly one y now the hard part :P is x a function of y?
No, cause each y gives us 2 x
It could be a function of yx
how so? elaborate ;)
|dw:1326051567892:dw| every element of a is maped to an element of b
y=x^2 yx=x Therefore each y gives us 2 x It is a function of yx
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?
y/x=x actually
Lol i rushed it
that's not solved for x, you have x on both sides :/
take the square root of both sides
y1/2=x?
is the 1/2 and exponent?
an exponent*
Yes, didnt you say that yesterday?
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...
ALright
plus or minus...
What?
when you solve\[x^2=4\]there are two answers, remember? one positive and one negative. or did you not know that?
No, why?
TURNING TEST BUDDY :D
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.
hi pokemon :)
ooooh
since negative multiplied by negative is positive
exactly :D
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...
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 ;)
so that said, back on topic\[r^2=t\]is t a function of r? is r a function of t?
r is a function of t but not vice versa
actually vice versa, but not vice versa lol
which is a function of what? careful.
So t is a function of r?
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
Ohh
each positive value of t will give two real answers*
Yeah, throw me another one
what about\[r=\sqrt t\]???
sqrt(t) cant be the dependent variable
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.
Oh, so if the +/- sign is gone then it can me the other way around?
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
yes because we are only looking at the positive root I'll draw...
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)...
Actually first the graph of r^2=t:|dw:1326054064202:dw|
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.
|dw:1326054308550:dw|^not a function|dw:1326054323340:dw|^is a function
because of the plus/minus it can be |dw:1326054372813:dw| or |dw:1326054351716:dw| Depending on if its negative or positive
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...
So that was wrong?
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?
No its not
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
ut its x, not y. Shouldnt it be horizontal line test?
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.
so another way to see it, is|dw:1326055195330:dw|is y a function of x?
Yes
good is x a function of y?
no
show please
|dw:1326055271855:dw|
perfect :)
:D
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 ?
y is a function of x and x is a function of y
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...
r | t --- 0 | 1 2 | 3 5 | 9 7 | 3 4 | -14 is r(t) a function? is t(r) a function?
No Yes
correct! very nice, I think you understand the concept of functions :D
Seriously? :D
yes, you are right :D care to state your rationale?
Rationale?
why? your reason.
how did you know?
Oh, for the answer?
yeah r | t --- 0 | 1 2 | 3 5 | 9 7 | 3 4 | -14 is r(t) a function? is t(r) a function WHY?
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
right, so r(t) does not represent a function, where t(r) does perfect!!!!
Yeah!!! :DDD
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 :)
Ok :D Im going to take a 30min break now :)
well deserved, take your time :D

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