#precalculus
1 messages · Page 295 of 1
Syst3ms
And so what you did was say $G'(x) = F'(x)-F'(1)$
Syst3ms
one final note
Sure
$F : x\mapsto \int_a^x f(t)dt+\lambda$ ($\lambda$ some constant) is a sort of "canonical" look for an antiderivative
Syst3ms
when you use lambda over c 
so, whenever you see something like this, you can immediately go "ah, F's an antiderivative of f, so F'=f$
Okok
conventions, conventions
I think I get it
Whenever you're not sure, take an antiderivative and write the integral in terms of it
Ok
Thanks once again
np
Just to verify if I got it, would F'(x) be x^2 . g(x) in the pic I gave?
correct
And if the upper and lower bounds of an integration is the same it gives 0 right? 
Since the area at a single point will be the area of a line which is 0?
yup
Thanks for bearing with me, helped me a lot
have you tried using the quadratic formula? (if you can't right away, what's stopping you)
yeah i have
i switched the sec squared to 1+tan squared and put it in the quadratic formula
idk what im doing wrong tho it keeps saying im wrong
did you consider the interval when taking the square root?
i didnt take a square root
how did you use the quadratic formula ?
there's a portion thats +- sqrt(b^2 - 4ac)
and the plus minus is an issue with trig stuff
so you could easily end up in the wrong quadrant
if one doesn't think about it
but also, you get multiple answers if that's what you want
depending on context
yeah, i see
and if it's electronic you might have to account for periodicity
i don't know what they want you to put
just the solution set
oh no that's just the +2πn
some systems want that
some don't
i think ?
not sure lol
im pretty sure i got it now thanks bro
can someone just tell me if they're all correct
Yeah, the points should be at the same location.
great thanks
Someone help evaluating this integral, exact answer not decimal and with full steps
What do you mean by "easier"?
Hm,
I'm not sure if this would be more helpful,
but I would solve this by writing down the first few primes (1, 2, 3, 5, 7, 11, 13) and try to see which numbers are divisible by them off the bat
1 doesn't matter,
2 is quite obvious (even or uneven number),
each number and its properties
I suppose it's a little too similar,
let me see if I can maybe find something else
Not quite sure if there is another method, sorry
Write all the numbers as prime factors
So (7x5)^4 * (7x2)^6 * …
Then you’ll find that powers of numbers will cancel
I’m sure that’s what you were doing anyway though
So I’m not really sure of a shorter way
Im currently doing the factor tree for each number
these numbers are relatively small and have very few factors, you shouldn't need to resort to drawing a factor tree for these
expand it like you would
(a + b)(c - d)
$\frac{7e^{\frac{3\pi i}{5}}}{0.5e^{\frac{\pi i}{5}}}$
Mosh
apply power laws
The 9th question
The answers are given as B and C
I'm able to prove A wrong and B correct
But am not sure how to go about solving C and D
I think I can say that D is incorrect as f(x) is increasing in (0, infinity) and therefore has no upper limit
Here's what I have written so far
<@&286206848099549185>
Lets put C in simpler words
For sufficiently large x, the derivative of the function is always smaller than the function
ok
magnitudes right?
oh wait f'(x) is always positive
im still not getting any ideas 
wait
@quick bluff
bro this question
is more like u cant probably prove it with absolute method in exam
but there is a more theoritical way
what is the "theoretical way"?
in the first derivative?
it wont affect much when we take higher values
or everywhere?
in every thing
ok 
we basically have to prove f(x)>f'(x)
how did you come to this?
look at part C
yes, cant it also be possible that f(x) is negative but of a greater magnitude?
if alpha>1 that means x>1 and modulus will open with a positve sign
oh
oh yeah
f(x) can't be negative then
now if it depends on lnx and 1/x
just look at the rates of increase or decrease of them
makes sense
and for the d part i proved it wrong
im still skeptical about the "sine term removal" you did tho
nah when terms get lager they become insignificant
the sine term integral cannot be more than 3 or 4
well can't i just say f(x) is always increasing in (0, infinity) and therefore it's not possible for it to have an upper limit?
it is
but the problem lies in the values like 20 to 80
mb there f'(x) goes up and comes down again
we dont know
and u can think of it from other methods
too
ok
im not sayin that my method is the only best method
if it's more concrete than my completely theoretical explanation it must be better
ya i just thought of it while havin lunch
i did not write it on paper
you're in allen??
@quick bluff
Yes
ask ur teacher and also tell me afterwards
Sure
Could someone help me out with this?
you can take constants out during differentiation. here you have a constant factor of 1/2
so d/dx (3x-1)^(-1/2)/2 =
(1/2) d/dx (3x-1)^(-1/2)
no problemo
<@&268886789983436800> why is math so hard aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
1951 4000000
1952 6000000
1953 9000000
1954 12000000
1955 16000000
1956 21000000
1957 26000000
1958 32000000
1959 39000000
1960 47000000
1961 56000000
1962 67000000
1963 80000000
1964 95000000
1965 112000000
1966 132000000
1967 155000000
1968 182000000
1969 214000000
1970 249000000
1971 287000000
1972 331000000
1973 382000000
1974 434000000
1975 480000000
1976 534000000
1977 593000000
1978 657000000
1979 728000000
1980 798000000
1981 870000000
1982 943000000
1983 1023000000
1984 1109000000
1985 1199000000
1986 1295000000
1987 1399000000
1988 1509000000
1989 1623000000
1990 1743000000
1991 1867000000
1992 1999000000
1993 2136000000
1994 2287000000
1995 2443000000
1996 2611000000
1997 2791000000
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2005 4603000000
2006 4883000000
2007 5178000000
2008 5459000000
2009 5747000000
2010 6060000000
2011 6385000000
2012 6723000000
2013 7075000000
2014 7442000000
2015 7823000000```
how do I derive a formula based on the above data set?
i ended up with f(x)=2.023*1.136^x, but my instructor said it was incorrect
i just followed a technique i found on a khan academy vid
how do u recommend i do it?
sry just pinging in case u dont even see this again lol
i mean it's just hard to discern how the dataset and the equation was derived
if you show what you did it's easier to point out where the misstep was
like where does 2.023 come form
i think it's worth revising what function you're trying to use f(x) = ab^x? some growth with a rate? how are you deriving your terms
and then think about it in simple terms
dont think of a huge dataset
what text are you using, it's always good to go back to that
which video did you use?
hey thanks for the response
that is the video I used
i followed the exact procedures as outlined in that video and ended up with the formula I have
the data set seems to show an exponential function of some sort, which is why i tried to use f(x)=ab^x form
do you mean the instructions of the assignment im working on?
it basically just asks us to find some sort of data set online that is a function of degree 3 or higher, or an exponential function, or trig function
and then to formulate an equation based on the data set that roughly models the data
500cos(30), 500sin(30)
anyone know how to find angle?
@inland flower apply the law of cosines
you could first consider the product of the constant components and eliminate some choices
yea but going from just the equation given
$4x^3-x^2-28x+7=x^2(4x-1)-7(4x-1)$
coycoy
how do u get that
factor out x^2 from the first two terms and -7 from the last two terms
just continue factoring after that
Jaba Teresinha
multiply the top by y and the bottom by -x. then add the two equations.
we cant answer, but we can help
sorry that's what I meant but I solved this already ty
Use D moivres
${(-1 + sqrt(5))/2}$
∂T/ ∂t = α ∇²T
did you mean $\frac{-1 + \sqrt{5}}{2}$
Ann
@slim echo divide all terms by 5
ok
I understand what you're saying but still unsure how this leads to my last term
does the n=20 just mean that whatever term I get when when I get to the 20th term will be my last term
nvm I was able to figure it out but thank you anyways
did i do this correctly
what kind of school activity is this?
this good?
you should not use the letter x for multiplication
and also you didn't do what was asked:
Write a full sentence to describe the recurrence relation of the sequence
$a_{n+1} = 3a_n + 1$ is what you meant for the first one
Ann
i had a sentence but i don't think i phrased it correctly
so i just ended up deleting it
Every value is 3 times greater, plus one than the previous value.
idk how to approach this sum can anyone help me
You need to find $$\lim_{x \to 0} f(x)$$and set f(0) equal to that
Lunasong the Supergay
how to find tat
||hiding so no one sees, but do you know Lhospitals rule?|| @wary mulch
S im not good at it
That's probably the easiest way
OK
The only other way I can think of is a little ugly
wats tat
why not just make x=0 though?
ignore me, my bad
does that even end well here?
i'd probably approach it with the factorisations for a difference of two powers
Oh, I was doing that, then it didn't work put, but I was dumb and only did it for the one power lol
Yes, that's the other way, but I mean, it's hella messy with an eighth root
Which I why recommend ||LH||
I don't think that's right
You didn't use chain rule when differentiating
my teacher didnt teach differentiations yet
Oh, so you aren't supposed to use LH?
You can't use LH if you haven't done differentiation
ohhh ok
I guess you have to do this then
ok
Multiply the numerator and denominator by
$[(256-7x)^\frac78 + 2(256-7x)^\frac68 + 2^2 (256-7x)^\frac58 + 2^3 (256-7x)^\frac48) + 2^4 (256-7x)^\frac38 + 2^5 (256-7x)^\frac28 + 2^6 (256-7x)^\frac18 + 2^7][(5x+32)^\frac45 + 2 (5x+32)^\frac35 + 2^2 (5x+32)^\frac25 + 2^3 (5x+32)^\frac15 + 2^4]$
Lunasong the Supergay
Lol
u kiddin me
You don't have to multiply out manually
So this comes from factorization
$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \dots + x y^{n-2} + y^{n-1}$
Lunasong the Supergay
So when you multiply out the numerator with the first bracket, you will just get 2^8-(256-7x)
And in the denominator, you will just get (5x+32)-2^5
Then with the extra pair of brackets
You get 7x times second bracket / 5x times first bracket
Cancelling out the x's gets rid of the 0/0 issue
And then you can substitute in
this is 0
2^8 - 256 + 7x = 7x
5x+32-2^5 = 5x
Cancel out the x's
Then it won't be 0/0 anymore
👍
This will work in general when you have 0/0 with roots like this
It just gets uglier the higher the roots are
8 and 5 pretty bad already lol
how did u findout with wat to multiply and divide so fast
factorisation identities
ohhh ok
This
I treat the one term as x, and the other as y
Then what I need to multiply with is that longass second bracket
And do it for the numerator and denominator
ok
could use sigma notation to avoid tedious writing
But then it doesn't look as crazy 
also makes evaluating the limit much easier in this case
I agreeeeee
old english font
can it be found on keyboard?
𝔸𝔹ℂ𝔻𝔼𝔽𝔾ℍ𝕀𝕁𝕂𝕃𝕄ℕ𝕆ℙℚℝ𝕊𝕋𝕌𝕍𝕎𝕏𝕐ℤ
that doesn't seem like its the entire question, might be helpful to post all of it
You dont need the equation
Just eyeball it
And reverse the coordinates of the given point
To find the inverse of the function, you swap the x and y variables right?
since you know its an exponential equation, the inverse would turn out to be logarithmic
honestly the easiest way is to just imagine what the graph would look like in your head when you swap the axis and then you can easily identify which ones must not be correct
if its easier, imagine the y=x line and mirror the graph to the other side of that line (someone else might be able to explain this better than i can)
wait y
eulers numbers?
e
this graph is exponential, right?
yea
so then whats the problem?
do u understand this part?
ok, now lets go through the options they provide
the first option is linear, is this a possibility for the graph u have visualised?
yea, it cant be it, cuz it should be exponential
and the second and fourth options cant be it, becaue x^2 should have a local extrema
ok, second and fourth option are quadratics, can they be a possibility?
yea
ok third is logarithmic, does the function u are visualizing look logarithimic to you?
yea ig
does the point they outline fit on the third option's graph?
0,2*
yes
2,0 exists for lnx/2
then u have your answer
is there any way you can get that without looking at multiple choice
ah, i see
you can prove it to yourself that one point is not enough to get an exponential equation, can u think of multiple exponential equations that go through that same point? use desmos graph tool and test around with some numbers
Yea will do
I spent a lot time trying to find equation for the graph, looking online
Couldn’t find anything
quick example
Yep
Acceleration is second derivative of x. Get that, find a(2)
When a scalar ,k is multiplied with a function f(x), we represent the resultant function as (kf)(x). But when me multiply f(x) with (k+m), where m is also a scalar, which is a better representation of the resultant function : ((k+m)f)(x) or (k+m)f ?
whichever you think looks better aesthetically
you could even say (k+m)f(x) since it's unambiguous what's being talked about
and likewise kf(x)
${\frac{x/x}}$
∂T/ ∂t = α ∇²T
${\frac{x/x}}$
```Compilation error:```! Argument of \frac has an extra }.
<inserted text>
\par
l.55 ${\frac{x/x}}
$
I've run across a `}' that doesn't seem to match anything.
For example, `\def\a#1{...}' and `\a}' would produce
this error. If you simply proceed now, the `\par' that
I've just inserted will cause me to report a runaway
argument that might be the root of the problem. But if
your `}' was spurious, just type `2' and it will go away.```
$\frac{x_x}$
suppose f is a real valued function, then kf(x) is a real number while (kf)(x) is a function ,right?
∂T/ ∂t = α ∇²T
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
@viscid thistle \frac{num}{denom}, also #bots or #latex-testing
no, kf(x) and (kf)(x) refer to the same thing. if you want just the function, it's kf.
@viscid thistle there is a cheat sheet pinned in #latex-testing if you want to familiarize yourself with the basics of latex for using the bot with
ohh k ty
While proving set of real valued functions is a vector spaces, kf(x) and (kf)(x) do not refer to the same thing. Maybe I should try asking this doubt in linear algebra channel. I thought this doubt was related to pre-calculus.
(kf)(x) and kf(x) both refer to a number tho
that is if you refuse to acknowledge that the former is defined as the latter
Write an equation for the parabola with focus (−1,2)and vertex (3,2).
for this problem, would I use the horizontal or vertical equation of a parabola
i think it's horizontal
try to think why though, try sketching it out
can anyone help me out with this question, no clue what to do here
You ca start by evaluating dy/dx
This will give the turning point
@viral juniper
okay cool
Hey, can someone help solve this?A rocket makes a one dimensional motion/movement.
g is the gravitational acceleration,
x is the vertical coordinate,
m is the mass,
t is the time.
Assuming that there is no air resistance/friction etc and only gravity affects the motion of the rocket, the force (F) is
F = –mg.
In this situation, the equation of motion (F= ma where a is acceleration) is (attached as a photo).
(1) Find and show the rocket's velocity as a function of time. However, when t = 0, its velocity is v0.
(2) Find and show the rocket’s position as a function of time. However, when t = 0, its position is x0.
(3) When v0 is much greater/a very large number, describe the rocket's motion.
How would you solve this problem? I'm thinking integrate once to get velocity, and twice to get position. It would be great if you could help out!
this doesn't belong here
don't post in multiple channels to get more attention
How does this fall into "pre"calculus loool
x = 1/2 at^2
v = at
Anyone able to help me figure out how to go about solving this?
What does the derivative tells you about a function?
When de function decrease and increase and the maxims and minimums
it wasnt an open question, it was a prompt
The instantaneous rate of change of the function
If you'd like I can help you with the question
it would be helpful if you posted the function itself but if you only want the knowledge, and then to apply it yourself thats better!
that... isnt the question
yep i didn't read it fully sorry
Could someone walk me through these? Going through my review and haven't done these in awhile
Synthetic division and stuff
I went back to try and solve it, I got this far with part a but I can’t remember how to go from here, or if I did something wrong
I used the rational zero theorem, do I have to finish it by using quadratic formula?
Alright I did quadratic formula and figured out another zero but I need one more, am I supposed to use that zero I just got and do synthetic division again?
Right, so if a derivative is negative you know how much is changing with each step right?
Yeah
So, there you know that if x<2 the derivative is negative, that means that with each step the function goes down right?
im getting a=1,b=0 but ans is a=1,b=1/2 can anyone help me
fully mind calculation
then do it on paper
mind calculation is cool for when you're walking without paper. it's almost always helpful to write some stuff out. even if it's easy, you might see some other properties that you wouldn't notice if it's all calculated by head
Wouldn't the answer be a > 0 and b = 0
Can anyone help me solve this problem?
Work out the modulus of the vector, which is 5
We want a vector with modulus 1, so our vector will be 1/5 of v given
yes I believe so
Why do you need to use modulus in calculating unit vector direction?
ahh too used to programming lol

why not just say magnitude or length, why modulus 😠
but yea that makes more sense now that I think about it lol
cause modulus is technically correct since C is iso to R^2, ie you can express complex numbers as tuples of real numbers
and modulus of a complex number is identical to norm of a vector in R^2
since both are just pythagorean
I see-ish
Still need help? What have you tried/where are you stuck
My doubt is if cos x = -1/2 , then x could be -60°
there are infinitely many solutions to cos(x) = -1/2
but -60° isn't one of them
anyone know how to graph this
like i dont know where to start/plot
do you know how to graph straight lines?
sort of
e.g would you be able to graph
y = x/2 + 4?
yes
start by doing that
seems that ← is supposed to be <
actually its supposed to be < -
and restrict the domain for that piece to (-inf, -4)
since f(x) only behaves that way for x < -4
so would graph this y = x/2 + 4
until where x is -4
since those are the conditions being given
f(x) = x/2 + 4 if x < -4
modulus is in fact normal usage for it
as mosh stated. also useful for dividing complex nums
Can commutative and associative laws over sum and product of integers be proved?
Is this an exam or some marked quiz?
no its a practice quiz for the finals its not graded
Hmm 5points is kind if sketchy
Anyway
Try to draw the tiles out
Draw the hexagon and squares out, fill them in
And see whats missing
And what kind of shape you can fill in that gap
im pretty sure i have to use the exterior angle of the green shape to get the number opf sides
cuz 360/the measure of one exterior angle = the number of sides in a figure
i think its 12nvm idk
Tho there is still one aspect of the question i dont quite understand, is the green shaded area supposed to be a whole shape?
Hmmm then why was the top corners coloured purple
It seems pike the square are supposed to be attaches to the top two sides of the hexagon
Thats where im kind of unclear about the problem
Tho but if the green shaded area is supposed to be a whole shape then yeah, 12 seems correct

@weak goblet
Define important variables. Hint: there's almost always 2.
Write the important equations - they give you a perimeter, so what's the equation for perimeter? They want an area, what's the equation for area?
Then, write area in terms of one variable. Differentiate. Set to 0.
I still dont understand, is it possible if you could do it for me? That just helps me better if I get a visual or something
Yes I do
are you ok with me taking you through this by having you do it step by step?
Yes please
okay, the first thing you should ask yourself is: what does the problem ask you to find?
i.e. what will the answer look like?
To find the area of the rectangle?
Oh okay
this should have been clear from a careful reading of the problem
your answer will look like:
The area of the rectangle is maximized when its length is ___ and its width is ___; this gives it an area of ___.
once we fill in the blanks for length and width, the area is (hopefully) easy to find.
therefore the quantities we care about are the length and width.
do you follow?
So we have to find the max length and max width?
no
it's area that is maximized. not either dimension on its own.
we didn't get there yet.
when solving math problems, typically you take the quantities you care about and assign them names, yes?
Yes
so why don't you do that now with the length and width?
which names would you like to call them?
L and W
alright
you should write at the beginning of your work:
L := length of the rectangle, in cm
W := width of the rectangle, in cm
Done ✅
do you understand why i'm having you write this?
no, it's so we can define our variables.
Thats what I was thinking lol
then you didn't word it properly.
Yea sorry so whats next?
anyway, ok, now we can get to the actual problem.
Okay
we care about the area of the rectangle.
so we will need to express it in terms of our variables.
How will we do that?
well do you know how to find the area of a rectangle?
length times width. don't use the letter x for multiplication.
Okay
but yes. so you just did exactly what was needed. why did you try to ask me?
What do you mean?
why did you ask me "How will we do that?" as if you didn't know?
I guess I just didnt think of it and I mightve been doubting myself
Yes
there is another piece of info we care about
the rectangle is enclosed by a border whose length is 36 cm.
Yes 36cm
look at the diagram to see the shape of the border, and translate the sentence "The length of the border is 36cm" into an equation.
L=36cm???
no
you're saying the length of the rectangle is 36 cm
that's not what you're told here
Oh the total length of 3 sides is 36cm
perhaps i should have said "The total length of the border is 36cm" so as to prevent you from getting confused.
yes, so translate that into an equation, will you?
Total length of border = 36cm
no. make that into an equation in L and W.
L•w=36cm
Yea
Length times width
I sorry im just really confused
which is not what we're talking about right now.
Oh okay
you yourself correctly responded with "L*W" when i asked you how to find the area of a rectangle
Yea I did
so then why do you say L*W again when prompted to express the total border length?
Oh right oops
try again.
Would it be a quadratic function?
you're overthinking it.
the words "quadratic function" here are extraneous, and should be ignored for now.
again
W^2 *L
width squared, times length???
I think so??
what kind of physical quantity do you get when you multiply three lengths together?
L^3
i'm looking for a word here.
Length cubed
that's called volume.
Yea
i asked you for the total border length and you reply with a volume?
does this not strike you as odd?
Yea im sorry again, so the total border is 36cm how would I write that as an equation
36
no, as an expression
L=36
36 = L*W^2
why are you insisting on squaring the width?
now you're adding an area to a length! this makes even less sense!
look. do you claim the perimeter of this triangle is (5 * 7 * 9) feet? because that's what it should be by your logic.
Im not sure what im supposed to do, sorry about that please help I will listen to what you say now
No
you should have listened to what i say from the very beginning. that you say you will do so now implies you didn't before, which is a bit disconcerting.
okay, so how do you find the perimeter of this triangle? of my triangle?
Yea It would be 9+7+5
yeah, so you would ADD the sides
not multiply them
not multiply one side by another and add the third
you would add them
now back to this
Yes I would so in my case, I would do width plus width plus length
2W + L
okay
great
so after all this blood, toil, tears and sweat, we have:
Area = LW
2W + L = 36
Goal: max Area
Yes!
the thing about our expression for area is that it has two variables, which makes it hard to maximize. maximizing functions of two or more variables is much harder than those of only one.
luckily, we have an equation relating our two variables.
this will let us isolate one of the variables, and plug the result into the expression for area. which in turn will make the expression for area contain one variable only.
do you understand this?
Yes I understand this part is clear to me
okay, great.
now would you be so kind as to follow up on this?
take our equation (2W + L = 36) and isolate one of the variables in it.
So like subtract either W or L
isolate either W or L.
Okay. Subtract L from both sides
so you've chosen to isolate W?
Yes
2W=36-L
What would I do now
Divide both sides by 2
yes
would have preferred if you had let shazz2005 figure this out for themselves, tenletters.
my apologies
W=18-2L
are you sure?
Or W=18-L
are you sure dividing 36-L by 2 gives you 18 - 2L? or 18 - L, for that matter?
why would the L miraculously stay intact (or, god forbid, get multiplied by 2?)
18-L/2
yes, so now you have W = 18 - L/2.
now follow up on this, will you?
Im not too sure how
well, now you have an expression for W in terms of L
Yeah
^
Ummm what am I supposed to do now @willow bear
this will let us isolate one of the variables [we did this already!], and plug the result into the expression for area.
Im still not to sure I understand
plug W = 18 - L/2 into Area = LW
W=18-L/2
?????
not quite
well now you just didn't do what i asked you
i did not ask you to just repeat the equation W = 18 - L/2.
Yea what do I plug in
i TOLD you!
i told you EXACTLY what to plug in, and what to plug it into!
i asked you to replace W with (18 - L/2) in Area = LW !!!
i can't understand what part of this is remotely unclear!
Oh okay so L*(18-L/2)
Area = L(18 - L/2)
yes
yes
finally
great
this took way more effort than it should've
but ok
so now you have that
Sorry
you have A = L(18 - L/2)
Yup
you can now change it to standard form if you want
this will make it more visible that it is in fact a quadratic in L, which is exactly what you were instructed to do
even though you did not need to know that, so the fact they said it explicitly should be treated only as a sanity check
what do you normally do with brackets?
Remove them
and how do you do that?
i suppose
okay then i'm out
Yes
and what do you get from that?
Do I distribute the L
Okay so it would be
Yes
do you know what do to get a maximum?
I do not
ah
Please help me out with that
you're going to need to get the derivative
Whats that and how
in a general sense, it is the instantaneous rate of change
as in, for a curve, it will give you the exact slope at any given point
you take the power of the variable, and multiply it to the front
and then take away one from the power
(in this case anyway)
So 18^2
not like that
but 18(2)
giving you 36
and when you derive a function, you usually write it as dy/dx
Oh okay
I would now have A=36L-L/2
you forgot something
Oh A=36L- L^2/2
L^1
not quite
36l^0
and what is anything to the zero power
1
therefore you get 36 for the first term
36-L^2/2
still got the second term to fix
remember:bring the power to the front and multiply it, and reduce the power by 1
36L^0
2-1=0?
No
2-1=1
yes
36-L^1/2
Ohhhh okay so 36-(2L^1)/2
yes
Okay
1
in order to find a maximum or a minimum, you equal the resulting derivative/slope function to zero
as when this occurs, the gradient/slope at the point is zero
Yea it is
Make one side 0
I have A=36-L
not a = 36-l
Im not sure…. Use the quadratic formula?
rather A'(L) = 36 - L
Okay
to indicate it's the derived function
well, just solve for your unknown
and why involve the quadratic formula?
Idk
36
Yes
Then what next
well, we had two variables to begin with
or, you can check if a function is maximum or minimum
by taking it's second derivative/ or drawing the graph
(of the original function)
the question is about max area is it not?
Yea we have to find the max area
well, in this case, since the highest power of the original function is 2
(therefore making it a quadratic)
there is only 1, turning point
and therefore only one value we can attain
Yea
if you take the second derivative, the resulting function will be simply 1
as all constants go away
@willow bear told me to write my answer like this. How would I
The area of the rectangle is maximized when its length is ___ and its width is ___; this gives it an area of ___.
as if you do the min/max test using the second derivative
it is greater than 1
which means it's technically a minimum
but lesser than zero is a maximum
regardless
you now have your value of L
36??
yes
Then what is w
but the original expression was 2W + L = 36
Yea
2*0+36=36
oh no
i just realized what went wrong
my apologies
we were working with 36 this whole time
it should've been 18
So my length is 18
and if we do take the second derivative (properly this time) we get -1
which is lesser than zero
and therefore a maximum
okok this is better
So what is my length?
Correct than width is 9?
Okay
write it like this now
i suppose
The area of the rectangle is maximized when its length is 18 and its width is 9 this gives it an area of
18*9
as for justifying it is a maximum area
you could say if the graph of the original function is drawn
it is a local maximum due to being a negative parabola
👍🏼
which the turning point of the graph (where the gradient of the graph is zero) is the highest point
or you could say you took the second derivative, which simply gave you -1
and since -1 is lesser than zero
But I cant have a negative width or length
Ahhh okay
yea that's about it
Thank you so much for your help today 😊😊 I have one more question but if you cant do it now tomorrow is fine as it is late for me
Ill leave it here and if you get a chance please look at it and tell me the steps I will do it tomorrow morning thanks for your help
why the answer is only A, why not A and E?
,rccw
oh okay , i got it, i need to place value for both sides.
i was just plugging negative value on left side.
mhm
given that this is true:
if log(u) = log(v), then u = v
does it follow that:
if log(u) = -(log(v)), then u = -v
is also true?
no
$\log(u)=\log(v)\iff u=v$
Mosh
$-\log(v)=\log(\frac{1}{v})$
Mosh
ooookay, thank you
hello one textbook I was reading stated that 6a9 is not a term, why is this?
<@&286206848099549185>
You're more likely to get help if you put effort into presenting the problem
Instead of sending a crummy photo
sussy
@sharp lagoon shouldn't it be set of all real no.s
f(x) is a polynomial
and all polynomials have a domain of R
ah ok ty
how would i write all real numbers in interval notation
$(-\infty ,\infty)$
Mosh
is dis algebra 1
LMAO i’m taking pre cal but that’s the “hw”
ohh LMAO
@sharp lagoon im pretty sure the domain is X element of R
r as in real numbers
okok!
Ok, this one I know the radical (x-3) would be x>_to 3 but what do i do with the 6 outside
I see,,,
what is wrong with my answer
well for one the domain is just R (or (-∞, +∞) if you so desire and/or are required to write)
but the range is not (-∞, +∞)
OO makes sense
OH i shouldve used desmo to graph it
i mean you didn't need to use desmos specifically but you would've done well to graph it yes
start by graphing it @tribal coral
then think about manipulating the expression ln(10x)e^(-x) into the inequality ln(10x) < e^(x)
it's completing the square
they're on the other side of the equation as well
if you're wondering why it's 4 and 1 specifically,
it's because x^2 - 4x can be turned into x^2 - 4x + (4 - 4) = (x^2 - 4x +4) - 4 = (x-2)^2 - 4
and similarly
y^2 + 2y = y^2 + 2y + (1 - 1) = (y^2 +2y +1) - 1 = (y + 1)^2 - 1
Thankyouuu^^
isn't it x >=3 ?
that's.. what they wrote
