#Can someone help me graph this one?
126 messages · Page 1 of 1 (latest)
Ohh sorry, I'm just trying to edit my text but can't do it, so i thought of just deleting it and making a new post
already it looks like theres a typo, this should be forall a, not forall x
Ohh yeah
needless to say theres a lot of requirements scattered around here
now youre allowed to use a piecewise function, right
Wdym?
have you learned what piecewise functions are
,,\begin{cases}-1&\text{if }x<0\1&\text{if }x\ge0\end{cases}
mtt
they look like this
Yeah, i know what piecewise are
alr, so we'll have to use some piecewises for each particular section
what we're looking for is just any easy function that can do whats required for each one
then the piecewise format can put them all together
first, we can consider (-inf, -1)
try coming up with a function where
- as x -> -inf, the function -> inf
- the function is differentiable
- the function is decreasing
the function needs to at least do that in (-inf, -1), the piecewise format will only look there
ignore the problem for now
just tell me a function that does this
Function? U mean make one to satisfy that? Is that necessary? I essentially tasked to just make a one graph given those conditions, I'm just struggling where to start.
cmon you gotta place some trust in me
and that begins by telling me an example of a function that does this, I can promise you it is useful for later on
you gotta communicate if you dont want to do it this way
do you want to do something else?
Sorry sorry, im just not quite sure how to do it
here, lets try something
lets say I give you three requirements, and you tell me an example of a function that fits those three requirements
I'll try
its separate from the problem for now, alright?
here are the three requirements
- as x -> -inf, the function -> inf
- the function is differentiable
- the function is decreasing
do you need help with any of these requirements?
I think a negative cubic
Ohh yeah
yea
thats all we need out of a function like this, straight lines will work for most of these sections
now heres the next set of requirements
- differentiable
- as x -> -1, f(x) -> 1
- as x -> -inf, f(x) -> inf
- f(0) = 0
- as x -> 0^-, f(x) -> 0
as you can see, theres more requirements
what function can do all of these requirements?
and again, dont overthink this
Is that a piecewise? I can't really think of any
have you tried y = -x a second time?
Oh crap
cmon man
Yeah mb
so already -x can do all of that
ok this time these are different requirements:
- continuous
- as x -> 0^+, f(x) -> 1
- f(1) = -1
- as x -> 1^-, f(x) -> -1
as before, try doing something simple like y = mx + b
-2x+1?
yep, thats what I got
heres another:
- continuous
- f(1) = -1
- as x -> 1^+, f(x) -> -1
- as x -> 2^-, f(x) -> 0
(this is not -x and not -2x + 1 btw)
Is this still simple function? What i can only think of is just a piecewise of x - 2 between 1 and 2, and 0 for >=2
the function y = x - 2 satisfies these requirements
do you understand that
I thought im supposed to find another function, im fogging my brain up, sorry
well you are finding another function
that function for example is y = x - 2
instead of having to be "y = x - 2 for 1 <= x < 2 and y = 0 for x >= 2"
again you can put your trust in me
you dont have to do the entire piecewise part by yourself, otherwise you wouldve done that by now
one step by a time alright?
Yeah sure
now this is a bit harder to pull off
try these requirements:
- continuous for x not equal to 3 (to make things easier to consider)
- as x -> 2^+, f(x) -> 0
- as x -> 3^-, f(x) -> inf
a line wont work for this one, but theres still many things that can fit these requirements
Idk, probably 1/3-x?
at x = 2, 1/(3-2) would be 1 instead of 0
theres a simple change you can do to 1/(3-x) to have it work out
Have numerator to be x-2? Or I'm overthinking this
thats interesting, I hadnt thought of that
that also works, its just as simple
I was thinking to do 1/(3 - x) - 1
that and this should be the same function
so (x-2)/(3-x) works for these requirements
alr last set of requirements
- continuous for x >= 3 (this can still mean its continuous for x >= 2, x >= 1, or whatever, it just at least has to be continuous for x >= 3)
- f(3) = 0
- as x -> inf, f(x) -> 2
Oh, 2(x-3)/x-2?
thats interesting, but you dont need the (x - 2) in the bottom, just x works
remember to put in parentheses
Ohh yeah
alr, so we now have a big pile of functions that satisfy some but not all of the requirements
Sure
of all the requirements that we have,
these are the functions that we ended up placing
all thats left is to figure out a piecewise that can be the correct function for the correct values of x
now heres the piecewise that we're working with:
[f(x)=\begin{cases}&\text{if }x<-1\&\text{if }x=-1\&\text{if }-1<x\le0\&\text{if }0<x\le1\&\text{if }1<x<2\&\text{if }2<x<3\&\text{if }x\ge3\end{cases}]
mtt
the idea here is that, in each case of the piecewise (each row), we write down a function, so that f(x) should fit every requirement
now conveniently Ive already sorted the requirements by interval then asked you about how to fill each one of them in
so for example, what function would go up here
Isn't that just the -x
yep
[f(x)=\begin{cases}-x&\text{if }x<-1\&\text{if }x=-1\&\text{if }{-1}<x\le0\&\text{if }0<x\le1\&\text{if }1<x<2\&\text{if }2<x<3\&\text{if }x\ge3\end{cases}]
mtt
now knowing that, try filling in the rest of the piecewise
if youre ever unsure, you can look back to the requirements to check that the function works/doesn't work
I'll try, brb
alr, gl
I just realized it's already in here
thats correct
technically, thats what youve already done so far, except for the dot colors and the 2 in the upper-left corner
So with the top conditions, some are just some removable discons?
I was also tasked to list the x's discons, where it's not diff., interval on where they are diff., and domains
I'll try to answer it and I'll let you check it
back
if you go for piecewise, you dont have to worry about removable discontinuities, since youre just manually writing down what each portion should/shouldn't do
removable discontinuities in this context would mean doing a (x-2)/(x-2) or something to prevent a function from being valid at 2 but changing nothing else
since we're using piecewises, we dont have to resort to that
we could just redefine the function at x=2 to for example not exist
you can see we have something like that for x = -1
Ohh i see, i just wanna ask if the domain would be from -inf, inf? I cant see any restrictions
Domain of the whole function
the domain is every number except 2, since the function needs f(2) to be DNE
Ohh yeah the asymptote. Btw thanks man