Hmm
#precalculus
viscid thistle
Oh wait
@alpine basin 15 and 75 is what we got together
But on degrees
Look
,ask 15° to radians
obsidian monolithBOT
viscid thistle
,ask 75° to radians
obsidian monolithBOT
alpine basin
okay thanks
viscid thistle
Ill think about the other 2 tomorrow, np
Okay i had to add that the problem included values at 2π which means that we can add 2π π/6+2π (which is one full revolution) we get $2\theta=\frac{13π}{6}$ and $x=\frac{13π}{12}$ which was also an answer. We can do this bc the interval is (0,2π] includes the 2π so we can add a revolution
obsidian monolithBOT
Al𝟛dium:
viscid thistle
@alpine basin ill tag you tomorrow for the other 2 solutions if you are still with doubt.
Bc i gtg sleep
viscid thistle
@still quarry Need help?
still quarry
Yes Please
obsidian monolithBOT
leviosa:
viscid thistle
Same with the 1^-.
still quarry
It means when approaching from the right and approaching from the left i think
viscid thistle
Right.
What is the value of f(x), when x is approaching 1 from the left side.
Which would be the 1^- side.
still quarry
It looks to be 1
still quarry
But theres a hole there
viscid thistle
Ight, so that's the concept about limits.
You approach a certain number (in some cases from a certain side) but never actually get there.
Meaning, you're infinitely close (in some cases from a certain side) to that number without being there.
I mean, if they just wanted you to input an x and get a y, you have already learned that many years ago through functions :/
still quarry
I get that stuff, but the point below
viscid thistle
Right.
For a limit to exist:
$$ lim{x \rightarrow h^+} = lim{x \rightarrow h^-}$$
obsidian monolithBOT
leviosa:
still quarry
So the 1+ would be 2 right?
viscid thistle
Yes.
viscid thistle
For a limit to exist.
Yes.
It has to be equal approaching to both sides.
When limits don't specify a "side to approach from," you can't assume which side to work with.
Thus, for a limit to exist (that doesn't specify a side it's approaching on), it has to be equal approaching both sides.
still quarry
So would i consider that one undefined
viscid thistle
Yes.
The line could be continuous and have an open hole at let's say x = 2.
And the limit would exist at x=2, BECAUSE when "approaching" from both sides, they're equal.
Just keep that in mind.
still quarry
Oh ok
viscid thistle
Need anything else?
Just consider limits as getting infinitely close to a number without actually being that number.
still quarry
Well I need to know if the function f(x) is discontinuous for any number x
viscid thistle
Well, it would be discontinuous if the limit doesn't exist no?
If at any point x, in which the limit is different from both sides, you have what's called a jump discontinuity.
still quarry
So I wouldn't count when x=1 as discontinuity?
viscid thistle
The limit doesn't exist tho.
At every point x, the limit must exist.
Meaning, at every point x, the x^+ and the x^- has to be the same.
Otherwise the graph would be discontinuous.
still quarry
In other words there is no discontinuity
viscid thistle
There is.
You can see it with your own eyes.
;-;
For example, this is a graph of a parabola and a straight line.
Because the limit at any point exists.
This is discontinuous.
still quarry
😑
viscid thistle
Because when approaching from the left and right side, they're not equal.
Thus, making the limit undefined.
viscid thistle
Np.
still quarry
Oh one quick thing
viscid thistle
Mhm?
still quarry
Do you know the proper formula for average rate of change?
viscid thistle
As in a line?
still quarry
Here
viscid thistle
$\frac{y_2-y_1}{x_2-x_1}$ or $ \lim_{x \rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x}$
obsidian monolithBOT
leviosa:
still quarry
viscid thistle
When t=1 what does C=?
still quarry
Temperature
viscid thistle
No like.
When time is 1, what is the temperature.
(Plug in t.)
I'm looking at a.
still quarry
Well 27
still quarry
As in get the derivative?
viscid thistle
Yes.
still quarry
No
still quarry
41
viscid thistle
$(1,27) (4, 41)$
obsidian monolithBOT
leviosa:
obsidian monolithBOT
leviosa:
viscid thistle
The "rise over run" formula.
still quarry
So I dont have to deal with that horrible f(x+h) formula?
viscid thistle
Nope.
If you've never seen it before.
Or never covered it in class.
There's no need for you to use that.
still quarry
Hmm it's funny how easy math can become when taught properly
viscid thistle
Have you seen that formula before tho?
It's actually not as hard as you think to understand.
still quarry
The y1y2 x1x2 formula?
viscid thistle
The f(x+h).
I can teach it in like 2 seconds.
It's easy to understand.
I'll just do it anyways and you can read through it if you want.
still quarry
I have seen it but i dont get it. Doing entire calculus in 3 weeks so not much time for soaking it in
viscid thistle
Let's say we have a function of $f(x)=x^2$ and we were to find the slope of it. $\$
$\$
You have probs learned the $\frac{y_2-y_1}{x_2-x_1}$ formula. $\$
And what you can take from that is, using 2 points, you can find the average rate of change. $\$
$\$
$\frac{f(x+h)-f(x)}{(x+h)-x}$ is the exact same formula. $\$
$\$
The main premise of a function $f(x)=y$ is that you plug in an $x$ and you are returned with a $y$. $\$
$ \$
Let's say we have a point, $(x,f(x))$, and we wanted to get another point without knowing anything about what x or f(x) is exactly. $\$
(Note that $f(x)=y$) $\$
Thus, we bring in an h. $\$
$\$
Now let's say that $h=1 \$
$f(x)=y$ would yield a point of $(x,y)$. $\$
And $f(x+1)=y$ would yield a point of $(x+1, f(x+1))$ $\$
Now that you have your two points, what's left is to find the slope of them, which leads to: $\$
If we plug these 2 points into the line of $\frac{y_2-y_1}{x_2-x_1}$ we get: $\$
$\$
$$\frac{f(x+h)-f(x)}{(x+h)-h} = \frac{f(x+h)-f(x)}{h}$$
I use h=1 as an example, to ease yourself into it.
But in reality, h can be anything.
still quarry
I will have to keep this for sure thank you
viscid thistle
Np.
still quarry
Do you think you would know how to do question c at all? Is it like guess and check?
obsidian monolithBOT
leviosa:
viscid thistle
Okay fixed some things.
Uhh lemme look.
You could solve it using diffrentiation, but not sure how else otherwise.
So probably.
still quarry
Ok
viscid thistle
@viscid thistle that is a beautiful explanation of the difference quotient.
Ima save that
rare zephyr
Out of xPy and xCy formula, in what context would make one of the formula more "useful"
odd helm
how would i do this problem? what i did was add the components of the 2 vectors and got (541.55, 333.36) and did arctan(333.35/541.55) to get 58.38 but that didn’t work
remote veldt
@odd helm first of all, make sure that you've converted angles appropriately (from west of north to what the angle actually is). Second, you want to subtract the vectors
Real velocity + wind velocity = effective velocity
you have effective velocity and wind velocity, you want the wind velocity
narrow peak
aight ima need some help w this qn
viscid thistle
narrow peak
The variables $x$ and $y$ are such that when values of $\frac{1}{y}+\frac{1}{x}$ are plotted against $\frac{1}{x}$, a straight line with gradient m is obtained. it is given that $y=\frac{1}{6}$ when x=1 and that $y=\frac{1}{2}$ when $x=\frac{1}{2}$.\(i) Find the value of m.\(ii)Find the value of $x$ when $\frac{3}{y}+\frac{3}{x}=3.$\(iii)Express y in terms of x.
this took 10 years to type
i got lazy at the end
but
yeah i need help
willow bear
1/x plotted against 1/x?
narrow peak
oh my god
willow bear
yea thats gonna give you a straight line with gradient 1
narrow peak
SHHH nothing happened
obsidian monolithBOT
Hmm:
narrow peak
aha yes
willow bear
have you already written down "when the values of 1/y + 1/x are plotted against 1/x, a straight line with gradient m is obtained" as an equation
willow bear
$y^{-1} + x^{-1} = mx^{-1} + b$
obsidian monolithBOT
Ann:
willow bear
you have two pairs of (x,y) values which are known to satisfy this equation
(1, 1/6) and (1/2, 1/2)
willow bear
you get a system of linear equations in m and b
willow bear
do you not know how to solve those
willow bear
did you really need me here
willow bear
have you found m and b
willow bear
it will certainly help to find b
willow bear
ok you have your equation, $y^{-1} + x^{-1} = mx^{-1} + b$
obsidian monolithBOT
Ann:
narrow peak
yes
willow bear
and you have your other equation, $3y^{-1} + 3x^{-1} = 3$
obsidian monolithBOT
Ann:
willow bear
this is, surprise surprise, another system of equations
hard hornet
tawny nacelle
lofty prism
what does it mean to describe a function SYMBOLICALLY and NUMERICALLY
wide ocean
not sure what to do
answer, but when I did the equation it did not result to this
subtle blaze
anyone know how to get the value of pi?
fluid shore
Sure, that's pretty easy
You just call up Dominos or Pizza Hut and ask for the price of a pie
subtle blaze
😢
fluid shore
what do you mean by the value of pi
subtle blaze
got no idea...doing this pre-cal hw right now LOL
fluid shore
show me the question
subtle blaze
I got the answer by guessing XD
I think you divide 32 by 2 and square root it to get 4
fluid shore
?? well, what's the maximum value of sin(2*theta)?
subtle blaze
oh I got no idea XD...i've finished 99% of the hw and this is the only one on projectile motion
fluid shore
So, what is the maximum value that sin(x) can take, for example
fluid shore
Be serious.
subtle blaze
I only remember sin2x=2sinxcosx
fluid shore
,w graph y = sin(x)
obsidian monolithBOT
subtle blaze
dead ass I took this class last year... I'm just helping a friend
ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
1
fluid shore
Yes
so sin(2*theta) has a maximum value of 1
and it achieves that when 2*theta = pi/2
and therefore, when theta = pi/4
fluid shore
no lol
subtle blaze
XDDDD
tawny nacelle
fluid shore
❤️ Don't mind if I do
fluid shore
I'm the Asian Indiana
fluid shore
you're welcome
subtle blaze
@fluid shore aye yo so if Cos2theta's max=0 then costheta=0 because 0/2 is 0 right?
:DDDD
fluid shore
?
subtle blaze
instead of sin2theta i'm saying hypothetically it's cos2theta
then the max of cos2theta=0 then costheta will be zero right?
because 0/2 is 0
fluid shore
,w graph y = cos(2x)
obsidian monolithBOT
subtle blaze
i saw the graph---------- so costheta=0
tawny nacelle
deep moon
So that's the differentiation of sin^3x
And there are two cases, both answers are correct but different . What am i missing ?
viscid thistle
sin(x^3) or sin(x)^3?
right side is unambiguously sin(x)^3
left side should really be written sin(x^3)
these are not the same functions on the left and right
deep moon
Yeah that's what i am a little confused about , they are not the same question right ?
viscid thistle
yes, the function being differentiated on the left is not the same as the function being differentiated on the right
the function on the left is sin(x^3), the function on the right is sin^3(x) = sin(x)^3
stuck lark
sin^3(x) = sin(x)
stuck lark
resisting copswing is an offense itself
viscid thistle
stuck lark
i was going to yell at you for lack of appropriate parentheses but idk if that's your work
viscid thistle
to remove any shred of possible ambiguity, the function on the left is sin(x^3) and the right is (sin(x))^3
deep moon
Thanks , i actually saw a yt video where a guy was solving sin^3x like sinx^3 , that made me think like what the fuck did i miss
viscid thistle
sin^3 x always means (sin(x))^3
deep moon
Ok , back to square one . If that's how it has to be solved and i have done it too but didn't realize it until now that i had this confusion . If we solve sin(x)^3 by taking the x^3 as t solving sint and then differentiate t .
Ahh wait
deep moon
Yeah ik that
Im kinda confuse about my confusion now , so like give me a min . I don't know if i thinking straight
viscid thistle
confused about your confusion
if we take a step back and start again at the picture, im pretty sure your confusion stems from the poor notation / absence of parentheses
deep moon
Yes
And now i get it, the answer changes when we change the t , plz ask me if this doesn't makes sense
I need to be in same page as you so you can understand my doubt
viscid thistle
to reference your picture specifically, $\sin x^3$ should be interpreted as $\sin(x^3)$, and $(\sin x)^3$ should be interpreted as $(\sin(x))^3$
obsidian monolithBOT
TTerra:
viscid thistle
what is t?
stuck lark
why did that font spread
viscid thistle
lmao i posted that and three people instantly started typing
deep moon
So like we have a question (sin x)^3 , if we replace the sinx with t the question would become t^3 , where the answer would be ( 3t^2 )(dt/dx), now we find dt/dx which would be cosx , so the answer becomes 3Sinx^2Cosx
viscid thistle
probably
you can blindly apply the chain rule here
wait nvm
copswing me!
you're right
fucking parentheses
3(sin x)^2 cos x should be the answer
stuck lark
you had enough for today
viscid thistle
i literally cannot take any more parentheses going missing
stuck lark
but i WILL start copswinging further lack of appropriate parentheses, eg
3Sinx^2Cosx
deep moon
you're right
Im ?
viscid thistle
you were right until you omitted parentheses
i can't copswing it seems, so pandacop it is
stuck lark
use copswing sparingly 
viscid thistle
my inability to comprehend mathematics aside, @deep moon you really need to be careful with using parentheses. as you can see, omitting or misusing parentheses can lead to you writing confusing or incorrect statements. while it may be correct in your head, you need to include these things so that other people can understand your work and confirm it
this goes back to the first picture, assuming that's what you wrote (if that's someone else's writing, or if that's your writing but copied exactly from someone else), then you need to be more careful
that is how i will sum up the past half hour or so
deep moon
Omg i am dumbfuck , it's now a fact my brain can't function properly after 14 hrs of work
I understand now
It's all about parenthesis
viscid thistle
14 hours?
viscid thistle
i dont think tired math is a great idea
deep moon
I wasted your time on my dumb confusion , sorry about that
deep moon
i dont think tired math is a great idea
Yes , but im trying to correct my time table so i shouldn't sleep rn and wait like 6 or 7 more hours so that i can wake and sleep like normal people
Yes i did
viscid thistle
good
deep moon
Maths is the safest bet for me rn , i need to study physics but i don't think so reading is gonna help me rn at least maths keeps challenging my brain so i can stay up
Anyways thanks for helping man
harsh smelt
physics is just advanced word problems haha
tawny nacelle
harsh smelt
hi soap
tawny nacelle
sup vimes
harsh smelt
but you are given their arguments
(f+g)(2) means that i find sum of f and g at point x = 2
copper vigil
wide ocean
Good morning
stumped
y=3, x=2
in quadrant 2 or 3
cos2theta is trig identity though, but it can simplify into three different ones
uncut mulch
you shouldn't be making that assumption just because the given ratio is 3/2
you should first consider which quadrant you are in.
specifically in which quadrant is tan positive and cos negative?
wide ocean
ohh
uncut mulch
technically that's not actually even needed
wide ocean
quad 2
uncut mulch
tan is negative in quadrant 2
uncut mulch
determine the value of cos(theta) and/or sin(theta) with the aid of a triangle (pythagoras, and/or similar identities)
doesn't really matter which form of the double angle identity you choose
note that I'm providing a general outline. depending on the conditions, certain shortcuts can be applied.
in this case, the cosine identity involves the squaring of sin(theta) and/or cos(theta) so you don't even need to worry about the signs
lofty prism
please can i get help
what does it mean to define a function or describe it symbolically and numerically
fluid shore
What kind of answer are you looking for?
I can make it as technical as you want me to make it. How technical do you want it to be?
@lofty prism
viscid thistle
Technical from abhi means T E C H N I C A L + 3 pages essay
lofty prism
ah shit
fluid shore
I'll give you two explanations. Pick and choose whichever one you want.
\
So, suppose i was just considering a function on $\bR$. Then, I would denote that by $f: \bR \to \bR$. $f$ is the name of the function. Let me give you an example of something that is a function:
$$f(x) = x^2$$
This is a function on $\bR$. It takes one given input and spits out a unique output. Every input is mapped to a unique output by the rule given above.
\
Let me show you something that isn't a function:
$$x^2+y^2 = 1$$
Let $x = \frac{1}{2}$. Then, $y^2 = \frac{3}{4}$. But $y$ has two solutions in this case. So, $x$ has two possible outputs and that's not what we want with a function. Every possible input in the domain has to be mapped to a unique output in the codomain.
obsidian monolithBOT
Abhijeet Vats:
fluid shore
Now, I'll give you the technical one.
Let $A$ and $B$ be non-empty sets. Then, a function $f:A \to B$ is defined as follows; $f \subset A \times B$ such that:
$$\forall a \in A: \exists! b \in B: (a,b) \in f$$
So, this is totally-defined and well-defined.
obsidian monolithBOT
Abhijeet Vats:
wide ocean
determine the value of cos(theta) and/or sin(theta) with the aid of a triangle (pythagoras, and/or similar identities)
doesn't really matter which form of the double angle identity you choose
@uncut mulch thank you so much
yeah I got the answer now
fluid shore
thats symbolically right
Yes, you can think of it symbolically. I believe the intuition given to kids is that if you draw the graph of the function on the xy-plane and if you draw a vertical line anywhere, the vertical line should not intersect with the graph more than once.
fluid shore
So, for instance, you can look at the graph of $x^2+y^2 = 1$. That's just a circle centered on the origin, with radius 1. The point is that if you draw a vertical line, then that vertical line will intersect the graph more than once and that cannot happen with a function.
obsidian monolithBOT
Abhijeet Vats:
wide ocean
thanks guys
fluid shore
there isn't a "numerical" definition of a function
The one I gave above is the most general definition
It doesn't imply that anything has to be numerical.
fluid shore
🤷
Probably just want you to give an example of a function or something
and describe it in 4 different ways
fluid shore
You're welcome
viscid thistle
is this correct?
willow bear
no
barren hedge
$\frac{(2n-1)!}{2n+1}$
dire bridge
oh crap typo
obsidian monolithBOT
4D61736F6E:
barren hedge
Let us know if you need more insight, but you're going to use that basically.
dire bridge
I tried to do the same with (2n-1)! but it just doesn't work the same as (2x+1)!
barren hedge
Try to make the bottom like the top
(sorry if I'm delayed, heading to the dentist heh)
obsidian monolithBOT
Hozay:
dire bridge
oops x to n
barren hedge
yup!
dire bridge
Thank you!
median ivy
do all functions have both left and right end behavior?
viscid thistle
@viscid thistle Make a function.
You have 2 points.
Why not find the slope, then use point slope?
Like this @viscid thistle
viscid thistle
@viscid thistle Bad slope.
It says.
For every 1 increase, there's a 5 child decrease.
Wouldn't the slope, which is "rise over run" be 1/(-5)?
So the function would be R(x)= -1/5x +11
@viscid thistle
inner sand
Hey, so what does it mean to use values of x that have integral square root values in a table of values. What are values of x that have integral square root values? (you can ping me)
remote veldt
@inner sand integral square root just means that the square root is an integer.
inner sand
@remote veldt Yeah I see thanks I got my answer
ancient stirrup
you can also rewrite f(x)=2x^(-1/2)
and apply f'(x)=2nx^(n-1)
with n=-1/2
also you can solve by v=sqrt(x) , v'=1/2sqrt(x) , u=2 , u'=0 and replace in the formula
still quarry
Such as this?
ancient stirrup
thats right
still quarry
Thx
ancient stirrup
np
fleet yew
@still quarry what are you having trouble with
still quarry
Well, specifically right now its determining derivatives and I've got 4 questions to figure out
stuck lark
freshman's dream
willow bear
i thought circumfrence is 18+19 =37
what led you to conclude that?
and then you thought pi = 1 apparently
@wide ocean
and where did you get it from that 18 + 19 would give you the diameter?
wym "usually"
have you put any thought into what the setup might actually look like?
"a+d" here would give you the height of the highest point on the wheel, which is not what you are looking for.
willow bear
wrong logic and overcomplicated.
the radius here is simply the amplitude of your sinusoid.
ie 18
mint bobcat
i thought diameter is 18+19 =37
@wide ocean oh lmao i tried to find height at h(0) first and it gave me 37, i wonder whats the relationship between h(0) and the circle
opaque olive
ill learn squeeze thereom and try it again
opaque olive
oh so i know that it must be bigger than 1 and smaller than something that tends to 1??
opaque olive
what after that
obsidian monolithBOT
HoboSas:
opaque olive
how did you deduce that
viscid thistle
Their ratio is close to 1
So they must be very similar
Define small
There's no need to
Small enough
Sin(0.01) = 1.7453292•10^-4
opaque olive
$\lim_{\theta \rightarrow 0} (\frac{\theta}{\sin\theta}) = 1$
obsidian monolithBOT
Yes:
viscid thistle
k then
opaque olive
${how would i get an approx for} \sin\theta$
pale bison
pls use parenthesis
viscid thistle
opaque olive
:/
pale bison
By part i and equation given in part ii, we see (by applying limit on all three terms):
${\color{green}{\lim_{\theta\to0}\frac{1}{\cos(\theta)}}}>{\color{cyan}{\lim_{\theta\to0}\frac{\theta}{\sin(\theta)}}}>{\color{green}{\lim_{\theta\to0}1}}$
since the term highlighted in green both goes to 1, what can you deduce?
viscid thistle
Can you make the middle one red
pale bison
no, i'm a rebel
viscid thistle
I cannot see the middle one anymore
obsidian monolithBOT
Publius:
viscid thistle
Nice
pale bison
since the term highlighted in green both goes to 1, what can you deduce?
@opaque olive
fiery wave
What are the available colors in the color pack on LaTeX
pale bison
not many, but you can use
\usepackage[dvipsnames]{xcolor} for a good variety
frozen needle
saying that 1>1 is kind awkward
pale bison
i should've put geq
willow bear
sinusoids do not have radii.
viscid thistle
how do you simplify this?
Okay
would it be B^X^2/y ?
thats what i got
So let's start from $x^{a-b}=\frac{x^a}{x^b}$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
would it be B^X^2/y ?
@viscid thistle no
Apply what i said above
@viscid thistle hello there?
yeah
$B^((x^2)/y)$
obsidian monolithBOT
alfred:
obsidian monolithBOT
alfred:
viscid thistle
So let's start from $c^{a-b}=\frac{c^a}{c^b}$
obsidian monolithBOT
Al𝟛dium:
obsidian monolithBOT
Al𝟛dium:
viscid thistle
What did you got after doing this?
b^ lnb(x)^2 / lnb(y)
Almost
You forgot the beta as a base @viscid thistle
idk how to write that on discord
Use b as before lol
Ill just write
oh i thought you meant the b after ln
so that's all?
do i have to simplify
$\frac{\beta^{\ln_{\beta}(x²)}}{\beta^{\ln_{\beta}(y)}}$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
No thats not at all all
so its just x^2 / y
Yes
I was gonna explain it lol
But yeah
oh
that was easier than what i did thanks
yeah i understand i think
just cancel out B^lnb
Just using the fact that $u^{\ln_u(v)}=v$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
No cancelling here
ln_b is pretty awkward
@viscid thistle there is no cancelling, just the use of the fact above
oh k
If you want a proof of that just let me know, and np 
viscid thistle
Do I just write 500 as the time ?
idk how to do this i need some help
Okay so
@viscid thistle if you lose 70%, how many percentage do we have left?
30
0.3 better
yeah
So we know that $A(t)=0.3*A_0$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
So $0.3\cdot A_0=A_0(1/2)^{\frac{t}{p}}$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
The A_0 cancels out on the next step
And you know time, t is 500
alright
Yeah
Solve for p with the instructions i just gave you
wait so i just write 500?
for time
Yeah
k
Use log properties and some algebra and you are done
-500ln(2) / ln (0.3) ?
@alpine basin #geometry-and-trigonometry
$\ln(0.3)=\ln(0.5^{500/p}) \ ln(0.3)=\frac{500}{p}\ln(0.5) \ p=\frac{500\ln(0.5)}{\ln(0.3)}$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
@viscid thistle
yay
oh wait
nvm
What? @viscid thistle
is there supposed to be a negative
oh nvm
1/2 =0.5
Yeah
No negative
k thanks
Np
opaque olive
so what i tried was find an expression for AP and BP first
and then squared the fact AP=2BP into $AP^2 = 4BP^2$
obsidian monolithBOT
Yes:
opaque olive
is that the right way to approach it?
i didnt get the answer though
(no spoilers pls)
viscid thistle
and would continuously just be 50000(e^0.5) ?
hazy cloak
summer sierra
Can someone please help me understand this concept
uncut mulch
show work
nova trail
@uncut mulch can u help me something simple
mint bobcat
@still quarry i got that too but i think i cheated a bit cuz i used the derivative test on a precalc question
nova trail
I just wanna check if I got something correct
A point T on a segment with endpoints D(1, 4) and F(7, 1) partitions the segment in a 2:1 ratio. Find T. You must show all work to receive credit. (10 points)
I got (5,2) for this, unsure if I got it correct. I think I did however
uncut mulch
what are you unsure about?
show your work
rich flint
@still quarry the horizontal tangent will more than likely be the vertex of the quadratic equation. Meaning remember the equation x=-b/2a that is another equation you can use along with the points given. Hopefully that helps
nova trail
@uncut mulch mk heres a picture
does it make sense
my work
and is the answer correct?
uncut mulch
,rotate
obsidian monolithBOT
uncut mulch
several things aren't needed
uncut mulch
1 sec
ancient stirrup
@still quarry remember that the geometrical interpretation of the derivate of $f(x)$ is the slope in every point of the curve that means the horizontal tangent line that passes to the curve happens when the slope is 0 or $f'(x)=0$.
obsidian monolithBOT
Kũrũrũ:
uncut mulch
oh, actually misread some stuff.
it wasn't explicitly clear why you were choosing addition or subtraction in your work
and its also better to represent it on an xy-plane instead of on separate number lines
plain abyss
uncut mulch
know your powers of 2
mint bobcat
if (-1,8) has a horizontal tangent then the derivative of f(x), f prime (x), equals to 0 for f prime (-1)
thats what i did so its kind of cheating lmao
uncut mulch
viscid thistle
Hi
glad pasture
can someone help me with these i forgot how to do them
viscid thistle
What do you mean I forgot
glad pasture
like i have not done it in like a few months and even before i didnt have the best understanding
since i was learning this online
glad pasture
would the first one be 7/8 ?
glad pasture
i though i was just supposed to convent it to radian
glad pasture
what do you mean not needed ?
glad pasture
and then you find cos,sin,csc and tan ?
glad pasture
seems like i have quite a bit to relearn if i forgot the unit circle
viscid thistle
You'll notice that cos(315)=cos(45) and sin(315)=-sin(45)
thorn wigeon
elfin cradle
i'm doing an act math section and i don't know how to answer one question
The expression sin^2(theta)-4+cos^2(theta) is equivalent to:
is sin^2(theta) = 1 and cos^2(theta) = 1/2?
elfin cradle
i have no idea what i'm doing
acoustic harbor
send a picture
viscid thistle
The most common identity
elfin cradle
isn't that equal to 1?
acoustic harbor
yes
viscid thistle
Almost
acoustic harbor
no
viscid thistle
No
elfin cradle
wait
acoustic harbor
ur joking
viscid thistle
Recap
acoustic harbor
ur joking
viscid thistle
Lol
acoustic harbor
yes -3
viscid thistle
Yeah -3
viscid thistle
You failed a 1-4 substraction 3 times lol, don't worry
opaque olive
how am i supposed to find out the asymptote of this function ? $\color{green}{y =\frac{x^2}{x-1} = \frac{1}{x-1} + x + 1}$
obsidian monolithBOT
Yes:
rough hatch
yes
uncut mulch
parentheses
opaque olive
parentheses
@uncut mulch where
viscid thistle
On 1/(x-1)
So the function is $y=\frac{x²}{x-1}$? And you wanna find the asymptote? @opaque olive
obsidian monolithBOT
Al𝟛dium:
uncut mulch
if you're sketching you'd need to adjust the notation a bit
opaque olive
f(x) = y
viscid thistle
Correct. The domain is $(-\infty , 1)\cup(1, \infty)$
obsidian monolithBOT
Al𝟛dium:
uncut mulch
specifically as x→inf, 1/(x-1) → 0^+
to indicate that y approaches x+1 from above
similar idea for x→ -inf
uncut mulch
union
opaque olive
union
@uncut mulch i havnt done math for 3 years 😩
uncut mulch
combination of sets
here, that combination represents the set of all real numbers except for 1
opaque olive
ok so domain is all real numbers except 1?
uncut mulch
\setminus
obsidian monolithBOT
Al𝟛dium:
uncut mulch
ugh
viscid thistle
ok so domain is all real numbers except 1?
@opaque olive yeah
ugh
How tho
Its whatever, just another way of expressing it
opaque olive
ok so what i dont understand iss why iss the asymptote at y = x + 1
viscid thistle
?
uncut mulch
as x gets large, 1/(x+1) → 0;
y will behave like x+1
uncut mulch
well its defined for x=-1 and 0 so you can plug those ones in
as for what happens around 1, consider what happens as
x→ 1^-
x→1^+
viscid thistle
Did we already established that 1 is an asymptote or what? I lost track of the conversation
opaque olive
unfortunately i didnt understand what he meant 😩
I need to sketch $y = \frac{x^2}{x-1}$
obsidian monolithBOT
Yes:
opaque olive
firstly, i shall work out y and x intercepts right?
and then deriving and finding stationary points
viscid thistle
We want to see if on vertical asymptote, on the right side of x=1 it approaches -infinity or infinity, same with left side
Have you been taught limits?
opaque olive
i should know them
opaque olive
ok so as x tends to 1 from positive numbers it apporaches infinity id say
viscid thistle
Wowowo
opaque olive
lol
viscid thistle
Let me write it
$\lim_{x\to 1^+}{\frac{x²}{x-1}}$ means the limit of the function as x approaches 1 FROM the right side
obsidian monolithBOT
Al𝟛dium:
opaque olive
um
viscid thistle
Which means from values like 1.001, 1.00001
opaque olive
so the superscript of + means from positive numbers?
viscid thistle
No
From the right side doesn't mean positive numbers, means from the right of x=1. Like values of x like 1.0001, 1.0000001
Like values close 1 from the right
opaque olive
so numbers larger than 1 sorry
viscid thistle
And from the left would be 0.99999, 0.9999999
so numbers larger than 1 sorry
@opaque olive close to 1, from the right side
I can't say it clearer
From the right side doesn't mean positive numbers, means from the right of x=1. Like values of x like 1.0001, 1.0000001
@viscid thistle .
opaque olive
1.00001 is larger than 1 lol
viscid thistle
Yeah but close to 1
opaque olive
yeah
opaque olive
ok
obsidian monolithBOT
Al𝟛dium:
opaque olive
yes i agree
viscid thistle
And then this one $\lim_{x\to 1^-}{\frac{x²}{x-1}}$
obsidian monolithBOT
Al𝟛dium:
viscid thistle
Correct
opaque olive
whats next
viscid thistle
Curvy you know
opaque olive
what
opaque olive
i thought we established that the right side was going positive and left negative
viscid thistle
@opaque olive have you looked up the y-intercepts and x axis intercepts?
opaque olive
0,0
viscid thistle
i thought we established that the right side was going positive and left negative
@opaque olive yeah thats how its represented
Horribly but you get the idea
viscid thistle
How did you got that may i ask
viscid thistle
Oh yeah i missread
viscid thistle
Why not lol
opaque olive
cuz asymptote
viscid thistle
Asymptotes is at 1 not 0
opaque olive
one of its points is (0,0)
viscid thistle
All you have to do next is do the derivative and check for increasing and decreasing intervals
opaque olive
yeah
viscid thistle
Set the derivative to 0 remember
opaque olive
yeah
viscid thistle
And add the 1
opaque olive
2,4 and 0,0
viscid thistle
Wow.
opaque olive
are my turning points
ive got everything except well the symmetry line if thats what its called
viscid thistle
Hm?
opaque olive
i know that on both sides of the vertical asymptote it looks the same
same shape
but different y values of course
ygm?
or not lol
viscid thistle
2,4 and 0,0
@opaque olive just checked correct.
opaque olive
ok yay
viscid thistle
i know that on both sides of the vertical asymptote it looks the same
@opaque olive i really don't know what you mean lol
viscid thistle
I mean yeah but you shouldn't assume that and it doesn't help either
opaque olive
like?
opaque olive
is there something called an oblique asymptote?
viscid thistle
Hold up
opaque olive
i should need oblique asymptote if i recall correctly
viscid thistle
But we did already said it
opaque olive
y = x +1 ?
opaque olive
ok
and now my question is why is that the oblique asymptote
and why does it equal that as x tends to infinity
viscid thistle
Hold up a min
opaque olive
sure
obsidian monolithBOT
Al𝟛dium:
viscid thistle
Evaluate it
Know how to?
This is for the slope of the oblique asymptote
Always you have to do $\lim_{x\to \infty}{\frac{f(x)}{x}}$ on this type of functions to get the slope of the oblique asymptote
@opaque olive
viscid thistle
How can you not see its occupied?
opaque olive
um
viscid thistle
Literally a message a second ago..
upbeat bone
@viscid thistle #❓how-to-get-help
viscid thistle
um
@opaque olive damn could you express yourself
opaque olive
gimme a few sec just tryna read what u said lol
viscid thistle
Okok
obsidian monolithBOT
Yes:
viscid thistle
Yes it does become that
how do you know thats the oblique asymptote
@opaque olive we haven't even got there
opaque olive
okay sorry carry on
viscid thistle
Have you evaluated the limit
opaque olive
doing it now
opaque olive
it becomes zero i assume
viscid thistle
No.
viscid thistle
No it does not
viscid thistle
$\lim_{x\to \infty}{\frac{\frac{x²}{x-1}}{x}}=\lim_{x\to \infty}{\frac{x²}{x²-x}}$
opaque olive
Al𝟛dium:
@obsidian monolith i agree
viscid thistle
wdym slope of the oblique asymptote
@opaque olive the eqn is y=mx+b where m is the slope you should know this
viscid thistle
$\lim_{x\to \infty}{\frac{\frac{x²}{x-1}}{x}}=\lim_{x\to \infty}{\frac{x²}{x²-x}}$ now select the leading x with the highest degrees
obsidian monolithBOT
Al𝟛dium:
viscid thistle
You should know that x² is the leading
obsidian monolithBOT
Yes:
opaque olive
?