#help-49
1 messages · Page 289 of 1
how about you answer that
||(I see. I just suspected that the graph might be concave up)||
no
hm?
help
what is your thought process?
i suck at math is what im thinking
where do you think the graph decreases
everywhere
but what
this is correct
altho more famously written as R
the REAL numbers
I'm not sure which form your fill in the blank thing would accept
yeah so its always decreasing on R and never increases
you already figured the asym thing
what about the domain
actually that was a pretty silly thing to ask
but yeah sure
ok cool
this is where we're at
or "touches" it
is the inc/dec the x values?
the what?
its always x value?
yes...
ok nice
I think?
I mean- I'm study functions and I've never used y-repreesntation for intervals
ok
actually yeah, the domain corresponds to x values
the y-values thing is called the range
in fact, its asking for it!
we already discussed the pos and neg part
pos is the part above the x-axis
and the neg is the part below
so how do i write that
This algebra video tutorial provides a basic introduction into interval notation. It explains how to express the solution of an inequality using a number line and interval notation. Examples include compound inequalities. You should use the parentheses symbol next to an infinity symbol or for open circles on the number line. You should use b...
intervals. they're a big deal 🙂
you sure learned how to do them but maybe forgot
so I'd def recommand watching the video.
thats not what my teach has down on other questions
can you show me
the said "other questions"
these are intervals.
exactly whats in the video
do watch it 😔.
they will teach you how to represent hte "pos" and "neg"
and everything else really.
the -inf , inf is also an interval.
Union you mean
yeah that's other stuff you def learned about
its when you want to combine two intervals
do we want to do that here
ill get to that when i need it
so that vid shows the greater than and less than signs, forgot what those are called, and looks easy, but how am i supposed to find that out from our question here
what? the union part is not the main concept taught
your teacher used intervals
i already learned that last semester
yes but you dont remember :((
you see how your teacher represented the range here
(1 , inf)
thats an interval
ok
sigh
I can explain intervals myself
because they are what you need here
will you promise to stay focused?
ill try
alright
so this is a not-to-scale numberline
its basically all the numbers you can think of (real numbers)
you know it!
it extends to infinity to the right and negative infinity to the left
are you with me so far
so far yes
amazing
so
we'll call this number line the x axis (it represents x values i guess.)
ok
1
so like- here
yes
alright
now what if i wanted to have x equal to many many things at once
for example
if I had a brother and I dont remember how old he is
but I siad
"my brother is in between 3 and 5 years old"
you can interpert that my brother is
3
3,5
4,5
one of those
5
ok
so
if i wanted to represent my brother's age in terms of x
I'll have to say
my brother is in ebtween x=3 and x=5
ok
yes
its kinda hard to put a point on EVERY x-value my brother can be
like
my brother can be
3 and a half years old
3.75
3.599
4.512
like a ton of ages in between 3 and 5
in fact, there is an infinite amount of numbers between 3 and 5
yeah there is- a lot.
we CANNOT represent them the way I did here
I dont have enough points or time for the amount of x-values i want to put here
so mathematicans created this NEW way of representation.
called intervals!
intervals are a way we represent values that are just too much to write down
nice
so yeah
graphically
this is how we denote an interval;
with the brown filling denoting that
EVERY value between 3 and 5
can be my brother's age
can be my x values
and
on paper
the way we denote them...
differs for one country to another
I denote them with [ ] but you seem to be denoting them with ( )
im not sure how these work...
but yeah let me teach you how to read and type an interval
so
at last
I wanna say, every number between 3 and 5 could be my brother's age
it is always x values. I'm just miscommunicating 
the way we'd write this on paper is...
(3 , 5)
with the 3 on one side and the 5 on the other
i understand the basics of intervals
basically the** boundaries** of my x values
perfect
now
if i perhaps
wanted to say
I want my x values to be
in between 1 and 20
how would you write that downh
(1,20)
good now
yes
notice soemthing very important here
the SMALLER number is ALWAYS on the left.
ok
( 4, -2) doesnt make sense
we ALWAYS start with teh smaller number when it comes to intervals
alright
now lets make it a bit trickier
what If I wanted to say....
I want my x values to be bigger than 4.
think about it
how could we represent htis
(4, infinity)
GOOD
but now here coems the part i dont fucking now
what do you guys do when you DONT want to include the 4 for example
... you dont understand dont you
okay what If I said
a human being cant be 0 years old
and lets say they only live to 60 years old
if i wanted to write the interval for these x-values
ME
i would do this
]0 , 60]
the ] denotes that "my x value CANNOT be 0"
if i wanted to include 0 as a part of my x values
I would [0 , 60]
what
give me 10 mins
a good chunk. Now I'm trying to explain how intervals work
I dont understand how people ues ( ) instead of [ ] for intervals.
in NA, we use [a,b] to mean [a,b], and (a,b) to mean ]a, b[
and ofc infinity is never closed
do you know the difference between [ ] and ( )
and when each is used
bracket is when the number is included
and ( ) when numbers are excluded
alright so
I- thought ( ) was how we say included
bruh
yeah so
its fine
we cna just rephrase my previous example to be "my brother's age is inbetween 3 and 5"
which would still be (3 , 5)
in any case
yeah so
if x can equal to 2 and can be bigger than it
how would you write that
x is greater than or equal to 2
x > line under it 2
nice
yes
so ALL of these values
and whats before -3
notice, pink area is where the graph goes below the x axis
we dont want that yet
we want the x values that gets the graph to be above the x axis.
which are the valeus i put in black + the values that are before -3 that we cannot see
are you with me?
yes
axis
true
but
we need to repreesnt that
in itnerval form
write it in the form of ( smth , smth )
or include stuff if we need to.
hmm
not quite
it sure does start with - infinity
but does it end with 0?
the graph is still abnove the x axis even when x=1
and 2
and 2.1
and 2.2
so 0 is def not the end of it.
(-inf, 2.585]
bang
I dont think we include 2.585 tho
idk when to use that
at the exact point of 2.585
the graph-
I might eb mistaken...
but yeah
in all
decaying
increasing
negative
positive
all these stuff
we only use ( )
ok nice
anyways yeah you did amazing
lets go
(2.585, inf)
and there you go
this is where we are
range
i know range is y values
the range is the only execption of which we represent it with y values
oh-
amazing.
ok
read from up to down.
let me make myself clearer
give me a bit
(3, -inf)
lets go
y-intercept, can you do that yourself?
yes
now idk if you're supposed to approximat- actually, where is the y-intercept on the graph
just from eyeballing it i would say it is (0, 2.5)
i dont think it is
how
do we plug something in
yes
what
you see, y-intercepts are just where the graph touches the y axis. do you know the equation for the y-axis?
theres an equation for the y axis?
so i dont
represnets soemthing that is applicant on all points of the y axis
ill give examples
( 0 , 3)
( 0 ,-1)
(0 , -9485)
(0 , 28)
these are all points on the y axis
notice a pattern?
x is always 0
so plug 0 in
||nice guess.||
ok nince
do you know what f(1) means)
plug in 1 for x
let me calculate it
you can use the graph.
or yea i can use the graph
f(1) means the value of y when x=1
so 2
when x is approaching - inf what is y doing
can you show me your teacher's example
and when x in approachig inf what is y doing
do that
seems easy
yes.
oh I highkey lost focus and got lost where are we in the problem
yes
end behavior
its okaty we're done
very almost.
ohk alr
was this yes to my question
end behavior with interval notation right?
yes
I'm not sure if you use interval notation for it
example from different questio
so first one is 3 and second one is - inf
depends on the teacher 
for x appraoches -inf the value y approaches is 3
andddd you're done!
I didnt even do that much
you did amazing pudgy.
for -infnity try to see what happens as it the graph moves left and for +infnity try to see what happens as the graph moves right (I know you're done but for future reference, yk)
any more questions?
thats just one part of the quiz
well- solve each one and send them here
let people check your work
as for me...
I shall study 
great luck
thank you very much
you're welcome
oh- you dont need to!
keep your grade to yourself and fill yourself with pride
what is there to be prideful of
if you need any extra help you can bother me
this is where math gets fucking annoying with derivatives. you're almost there ❤️ (its not hard, its just an interesting process)
alright, goodbye 
why did teach write range is 1 , inf
alright so look at the graph
is inf>1?
yes
so teach is right?
yea
lowest always first
infnity is not less than 1 so it goes after got it?
I'll have to go offline for a bit but any questions you can just dm me and il try to get to them later @proper swift
Closed by @proper swift
Use .reopen if this was a mistake.
.end
neither?
isnt growing anmd isnt decarying
well yes
.open
type .reopen
.reopen
✅ Original question: #help-49 message
didnt I just tell you what the difference was
Its my fault, I didnt see that you told him.
its better here
pudgy this is why youd dont open multiple channels
I told him to reopen
you can go on tho.
its my first time
go read what I said then go back here
here, f constantly heads downward instead of upward, so it wouldnt look like "growth", that could be why
you can tell it goes downward because its -(4)^x instead of +(4^x)
the - would mean youre subtracting larger and larger numbers as x gets larger
this doesnt make sense
what about the picture
decay
not what I meant
ok nice
what function is this graphing
i dont know what that means
you know how you need a function to graph something?
yes
so what function did I use to get this graph?
and remember, this is what I said directly before that
thats good
now heres the big problem with "growth" or "decay"
mathematically we call things getting faster and steeper "growth"
while we call things that settle or more lenient "decay"
so in this graph of 5 functions, all the ones going up or going down Id call "growth"
because its growing, two of them are growing up, two of them are growing down
this is different than what your teacher seems to be doing
more complicated I think
it might just be that when a is negative it is neither
yep
but it might not be
in this case it should be safe, |b| > 1 and a < 0 should be "neither"
what
you said "when a is negative", so youre using f(x) = ab^x in general
if |b| > 1 and a < 0, then that should mean "neither" to your teacher
I thought
neither
meant
for when f'(x) = 0
@twilit jetty
the slope of the tangent when its horizontal
car, as I said earlier, the teacher is using a different, more complicated definition of growth/decay/neither
so we have to go with whatever the teacher is cooking
also, this is not true
sin(x) does not grow or decay as x -> ∞
sin(x)'s derivative though is cos(x), not 0
I meant for whatever values makes f'(x) = 0 aka when the slope of the tangent line is 0 at some point, at that specific point, the function doesnt increase or decrease 😟
this is also not true, growth is not the same thing as increasing
this would mean sin(x) is "growing" when it goes up and "decaying" when it goes down
oh noo 😭
I was- helping this person earlier....
I interperted decaying as decreasing and increasing as growing
growth/decay as math words are usually taught along exponentials, if they dont teach negative a then its easy to think thats what the words mean
the function I described to him was an exponential one
give me a minute
oh... wait the growth part is out of the- okay nevermind
sorry.
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heyo i need to make a proof of product rule and ill show some process I did
```latex
\begin{align*}
p'(x) &= \lim_{h \to 0}\frac{p(x+h)-p(x)}{h}\
&=\lim_{h \to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\
\end{align*}
# real car fan
what I can do after this?
I think something to try is expand out f'g + fg' with the limit definition
then see how you can maybe transform what you have into that
do what he said
can I just write $-f(x)g(x+h)+f(x)g(x+h)$?
# real car fan
Yes that's a good idea
go on
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Hello, could someone check if this proof looks good please?
```latex
\begin{Theorem}
Assume $a,b \in \bR$. If $a \in \bQ$ and $ab \notin \bQ$, then $b \notin \bQ$.
\end{Theorem}
\begin{proof}
Assume for a contradiction that $a \in \bQ$, $ab \notin \bQ$, and $b \in \bQ$.
Since $a$ and $b$ are rationals, we can express both as
$a = \frac{m}{n}$ and $b = \frac{p}{a}$
where $p,q,m$ and $n$ are integers.
Then, $$ab = \frac{m}{n} \cdot \frac{p}{q} = \frac{mp}{nq}$$
Since $mn$ and $pq$ are integers, $\frac{mp}{nq}$ is rational.
This is a contradiction, since we assumed that $ab$ is irrational.
Therefore, if $a \in \bQ$ and $b \notin \bQ$, then $b \notin \bQ$.
\end{proof}
Mor Bras
looks fine aside from the typo in the last line
also b=p/q
u wrote b=p/a
we need mp and nq to be integers, not mn and pq
and nq nonzero
@fickle sierra Has your question been resolved?
Here's the the fix
```latex
\begin{proof}
Assume for a contradiction that $a \in \bQ$, $ab \notin \bQ$, and $b \in \bQ$.
Since $a$ and $b$ are rationals, we can express both as
$a = \frac{m}{n}$ and $b = \frac{p}{q}$
where $p$ and $m$ are integers, and $n$ and $q$ non zero integers.
Then, $$ab = \frac{m}{n} \cdot \frac{p}{q} = \frac{mp}{nq}$$
Since $mp$ are integers and $nq$ non zero integers, $\frac{mp}{nq}$ is rational.
This is a contradiction, since we assumed that $ab$ is irrational.
Therefore, if $a \in \bQ$ and $ab \notin \bQ$, then $b \notin \bQ$.
\end{proof}
Mor Bras
Thanks for you comment!
.close
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Help for number 1 for slop and secant line equation
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Ok
,rccw
Well, to start with, m should be a coefficient, which doesnt seem to be the case here.
.. find and name the points through the secant properly first
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Let the projection map be $\phi$. Then $\phi(p(x))= p(x) +((f(x)))$
wai
I suppose I now have to determine what $((f(x))$ looks like
wai