precalculus

[uncut mulch]

for this question you should multiply by the conjugate of the numerator

[viscid thistle]

^

[lethal oracle]

Why is that better in this case?

[viscid thistle]

Because cos^2-sin^2=1-2sin^2 which is what you're looking for.

[lethal oracle]

Alright

[viscid thistle]

Because the numerator of what you're trying to get is -(1-2sin^2).

Use trig identities.

To help you.

This proving business is basically a test on if you know your trig identities and can you use it.

[winter isle]

how would i rewrite this as an exponential function: y = log(x-2) + 5

wouldn't it be 10^(y-5) = x - 2 ?

[viscid thistle]

What.

[lethal oracle]

Alright I made it to the answer

[viscid thistle]

$y-5=\log_{10}(x-2) \ 10^{y-5}=x-2 \ x=10^{y-5}+2$???

[obsidian monolith]

leviosa:

[viscid thistle]

@winter isle

[lethal oracle]

Thanks for the great kelp

Help*

[viscid thistle]

Np 👍

[lethal oracle]

I might have some more questions, I have an exam tomorrow

[winter isle]

yeah that's right

thank you

[viscid thistle]

Np 👍

[lethal oracle]

Is there a better way to do this other than just making the denominators the same and adding them

[viscid thistle]

Do that.

Just remember that cotxtanx=1.

[lethal oracle]

Do I substitute the terms in the numerator using Pythagorean identities?

[blazing raven]

I'd multiply the left fraction on the left side by cos x / cos x.
I'd multiply the right fraction on the left side by sin x/sin x.

@lethal oracle for 45.

[toxic jasper]

What

Lol

[blazing raven]

oops miscall 😦

[viscid thistle]

?

[leaden stratus]

How would you solve number 217?

[serene heath]

Try changing all bases into 2

[leaden stratus]

So 2^(3x) + (2^3)^x = 2^(1/5)?

[uncut mulch]

parentheses

[leaden stratus]

?

[uncut mulch]

2^(3x) + (2^3)^x = 2^(1/5)

[leaden stratus]

Oh, yes. That's what I meant.

Ok, edited. Thanks.

After this, how would you proceed? We can't sum the exponents, because it is a sum and 2^(1/5) hasn't got any x.

[willow bear]

2^(3x) and (2^3)^x are one and the same

so you have 2^(3x) * 2 = 2^(1/5)

[leaden stratus]

Wow! Thanks. Solved it.

[green zenith]

hello! i need some help solving either a or b in 1=a^(-1)+b

[serene heath]

What's stopping you there?

[green zenith]

i didnt post my full problem. my bad.

so if i solve for b, i wouldve got 1-a^(-1).

then i have to find the full answer with the second point in my paragraph (-2,7)

the function is f(x)=a^(x)+b

so i got a and i have my second point

im soory, i have b rather

so 7=a^(-2)+1-a^(-1)

[willow bear]

and now that's a quadratic in a^-1

[green zenith]

so does that mean i have 2 a's?

[willow bear]

actually... wait a minute

no nvm

[green zenith]

so a^(-1)-1 =7

[willow bear]

....what?

[green zenith]

then a^(-1)=6

[willow bear]

no?

[green zenith]

no?

[willow bear]

where'd you get a^(-1) - 1 = 7?

[green zenith]

well its (a^(-1)-1)^2=7?

[willow bear]

(a^(-1) - 1)^2 is not a^(-2) + 1 - a^(-1)

[green zenith]

oh yeah thats right

[willow bear]

anyway, you'll get two values for a but you'll want to discard the negative one

[green zenith]

so i can factor a^(-2) + 1 - a^(-1)

[willow bear]

you can try but it'll be of no use

the other side isn't zero

[grim current]

Try to recall the other way in which we can write numbers with negative exponents.

[green zenith]

okay

i still want to know the full solution so i dont get lost

1/a^2-1/a+1

how do i even factor this

can i use the quadratic formula on this one?

[grim current]

Where did the "1/(a+1)" come from though?

Also, the negative sign.

a^(-1) isn't the same as 1/(a+1)

[green zenith]

i just switched them up

i switched 1/a and +1

it is separate

[grim current]

Aah, never mind.

I thought you put "a+1" as the denominator.

[grim current]

You'll have to do the LCM for a^2 and a.

[obsidian monolith]

Tello:

[keen breach]

using logarithms

taking log of both sides you can drag the power down (property of logarithms) and do some algebraic manipulation to solve for x

[stuck lark]

since it's been spoiled already, guess i'll latex it out

[obsidian monolith]

RokettoJanpu:

$\log(a^b) = b \cdot \log(a)$

[keen breach]

also maybe making them the same power might help to allow for easier algberaic manipulation (you could also do this with the other side by making 5^(x-1))

[obsidian monolith]

arcsine:

Tello:

[errant timber]

unless i did something wrong

[keen breach]

yeah that's going to be hard to play around with so i recommend you convert them to the same power

then you can do a

$\frac{a^n}{b^n} = (\frac{a}{b})^n$

[obsidian monolith]

arcsine:

[reef rune]

what is the equation of g

I know that f is x^1/3 so for g I have -x^1/3 right now

I can't figure out the transformations of g though

[reef rune]

<@&286206848099549185>

[viscid thistle]

Hi Pingu.

[reef rune]

hi rudy

[viscid thistle]

@reef rune how do you know that x is ^1/3 huh

[reef rune]

well

it looks like it

I'm assuming it's 1/3 lmao

and I got it right on the test when I wrote 1/3

but honestly idk wtf the transformations are here

[viscid thistle]

Well

We know it ISN’T -x^(1/3)

Because it doesn’t...

[reef rune]

yeah

[viscid thistle]

It doesn’t what Pingu

[reef rune]

hit those points

[viscid thistle]

More or less, yeah.

[reef rune]

lol

what would you think g is

[reef rune]

$$\left(-x+\left(-4\right)\right)^{\frac{1}{3}}+2$$

[obsidian monolith]

Pingu:

[reef rune]

found it lol

[green zenith]

what is the equivalent of a^log a(x)=x in exponential form?

[hoary valley]

I chose the range is [-1, +infinity) , It says that my answer is wrong..Is it wrong??

[pale kettle]

You might be switching up the range and domain

[hoary valley]

What you mean ?

[pale kettle]

What does range mean?

[hoary valley]

I'm not switching

The output values of y

[pale kettle]

And what are they here

Can you get an output value of -1

[hoary valley]

yes

-1

[pale kettle]

where

where can you get a y value of -1

[hoary valley]

Wait at -1, I get 0

Right?

[pale kettle]

Yep

That's not a y value of -1

[hoary valley]

When x is -1 I get 0

So should be [0,+infinity)

Right?

[pale kettle]

Yep

[hoary valley]

Thank you man.. ❤ ❤ ❤

[viscid thistle]

@hoary valley answer is a

[bright lichen]

hi i’m sorry if this isn’t the right group to post this question in! but i need some help showing that these 2 functions are equivalent if anyone could help me

[west kindle]

have you learned what conjugates are?

[bright lichen]

no i haven’t came across that term

is that an A-math topic that is required to solve this question?

[prisma prairie]

how is this equal?

where does the 2 come from

[uncut mulch]

they are not equal

[noble halo]

So for one of the problems I’m doing i have to find the the interval(s) where the function is increasing and the answer that I’m seeing is from [0,1] U [3,4]

It’s number 3

And I’m just confused on why it’s that

[inland forum]

think of that graph as a roller coaster @noble halo

the roller coaster goes up from [0,1]

goes down

then goes back up from [3,4]

[ionic bolt]

it isn't increasing at 1 though, so make sure to use parentheses

[noble halo]

Thanks

[prisma prairie]

$is \ln \left(y-1\right) = \ln y-\ln$ ?

[obsidian monolith]

The-Elite:

[keen breach]

no

ln is an operator so you can't just multiply it like that

[prisma prairie]

ah ok

ty

[keen breach]

@prisma prairie i made a mistake lol wait

[prisma prairie]

yeah

?

[keen breach]

$\ln a - \ln b = \ln(\frac{a}{b})$

[obsidian monolith]

arcsine:

[keen breach]

so ln(y-1) is already the simplest form

[prisma prairie]

so ln(a-b) = ln(a/b)?

[keen breach]

no

ln a - ln b, i made a mistake just now

[prisma prairie]

so what does ln(a-b) = ?

@keen breach

[keen breach]

nothing

that's the simplest

[prisma prairie]

eah

yeah

but

whats before that

can we go backwards for a moment

what would simplify to ln(a-b)

@keen breach

[keen breach]

nothin

[prisma prairie]

what do you mean nothing

[keen breach]

nothing basic like

[prisma prairie]

whats the step before that

bruhhhh

[stuck lark]

perhaps ln(a-b+0)

[keen breach]

if you take the natural log of a-b then you'll get that

[prisma prairie]

😦

ok

so

if we have

e^{ln\left(y-1\right)}:=:y:-:1:?

$e^{ln\left(y-1\right)}:=:y:-:1:?$

[obsidian monolith]

The-Elite:

[prisma prairie]

is that right?

[stuck lark]

👍🏽

[prisma prairie]

ok thanks bro

[green zenith]

does evaluating logarithmic functions have solutions?

for example log 10,000?

it is 10,000=10^x

[tall granite]

what

[green zenith]

is there a step by step solution for solving x?

i know youre supposed to equate the bases so you can find x

but i want to know the step by step as to understand it

[heady jewel]

give an example of a question

equating bases

is logarithms

in another form

[tall granite]

just tinker with it until the answer pops out

[green zenith]

for example log 10,000

10,000=10^x

then 10^4=10^x

so x=4

is there any solution in finding 4?

or is that it?

[tall granite]

that's it

[green zenith]

oh

okay thanks

[heady jewel]

the base has to be 10 firstly @green zenith

and what you wrote is literally the definition of logarithms

$\log_{n} a=y$ implies $a=n^y$

[obsidian monolith]

Lionel:

[willow bear]

iff

[green zenith]

@heady jewel so after equating the bases, i just look for the exponents

[heady jewel]

yeah

[green zenith]

okay thank you

how do i determine my domain if i have two x's?

for example if i have (x-2)(2x-1)>0

i have x>2 or x>1/2?

[tall granite]

saying x>2 or x>1/2 is the same as saying x>1/2

whenever the sign of the function changes, it has to pass through a zero

so zeroes dictate the endpoints of the intervals with constant signs

all you need is to sample one point from each interval to know the sign

or do it by inspecting the factors without even sampling any points

[uncut mulch]

*incorrect use of the term domain

domain is where the function is defined
what you're referring to are the solutions to the inequality

[tall granite]

he's probably dealing with a rational function where the expression is the denominator

hence domain

or so I would assume...

[green zenith]

what im dealing with is the domain of logarithmic function log(x-2) + log(2x-1)

so my domain then (1/2,infinity)?

and again, im sorry for asking too much basic questions.

[uncut mulch]

in this case, it would be
x>2 and x>1/2

so it won't be (1/2, inf)

what's the solution to: x>2 and x>1/2?

[green zenith]

so x>2?

[uncut mulch]

yeh

do you understand why we're using and here?

[green zenith]

not really

[uncut mulch]

both log(x-2) and log(2x-1) need to be defined

so (x-2) > 0 and (2x-1) > 0

[green zenith]

the statement "and"

so it can only include what is in the both of them

[uncut mulch]

yeh

[green zenith]

so in rational functions, its or, and in logarithmic, its and?

[uncut mulch]

for domain its pretty much all and

[green zenith]

even for rational fucntions?

[uncut mulch]

since all parts need to be defined

yeh

[green zenith]

how about rational functions that have square roots?

if it was 1/sqrt((x-2)(2x-1))

[uncut mulch]

$\frac{1}{\sqrt{(x-2)(2x-1})}$

[obsidian monolith]

ramonov:

[green zenith]

would that be still be and?

or or?

[uncut mulch]

the square root is by itself so there isn't anything to use and with
but the inequality for the inside will be or
(x-2)(2x-1) > 0

btw

x>2 or x>1/2?
wasn't the correct solution to that inequality

[green zenith]

do you mean that is not the solution to the one inside the square root?

[uncut mulch]

not the solution to:
(x-2)(2x-1) > 0

[green zenith]

how come?

[uncut mulch]

(x-2)(2x-1) > 0 doesn't mean
(x-2) > 0 or (2x-1) > 0

[green zenith]

i'm confused

[uncut mulch]

i recommend plotting (x-2)(2x-1)

[green zenith]

you mean the f(x)=(x-2)(2x-1)?

[uncut mulch]

yeh

just a rough sketch will be enough

[green zenith]

wouldn't the domain be all R?

[uncut mulch]

that graph doesn't look right
where are your x intercepts?

[green zenith]

oh yeah youre right

[uncut mulch]

don't worry that much about specific points
focus on the 2 x-intercepts and the general shape

[uncut mulch]

from the graph, can you determine when (x-2)(2x-1)>0?

[green zenith]

when x=1/2?

and x=2

[uncut mulch]

f(x) = 0 at those points
when is f(x) > 0?

[green zenith]

when x<1/2 and x<2?

x>2*

[uncut mulch]

yeh

[green zenith]

that is the solution of 1/sqrt(x-2)(2x-1)?

the domain would be (-infinity,1/2)U(2,infinity)

so its neither and nor or?

[still yew]

Something isn't clear to me, so it tells me to identify the quadrant under 2 conditions.

The conditions are tan>0 and csc < 0

I am confused about csc<0

I know that csc = 1/sin.

But how does that influence anything with actual sin?

[willow bear]

the reciprocal of a positive number is again positive

the reciprocal of a negative number is again negative

[still yew]

Ah, thx.

[viscid thistle]

Hi all, im currently taking trigonometry, and have gotten confused with functions of it.

specifically graphing it.

my professor said to graph the function when lost

[flint stirrup]

any specific functions?

[viscid thistle]

im guessing basic

cos and sin

my prof. is garbage at teaching

sorry

[flint stirrup]

do you know what they look like?

[viscid thistle]

waves?

[flint stirrup]

pretty much

is there a problem you are solving?

we could work througuh it

[viscid thistle]

im looking through my hw right now. give me a sec.

and thank you

[flint stirrup]

🙂

[viscid thistle]

i do understand the transformations

for the most part

[flint stirrup]

i see

[viscid thistle]

like its shifted to the left by 4

and up 5

and the amp is 3

the midline im a little lost on

[flint stirrup]

alright so

lets look at each part individually

how does the 3 in front change it?

[viscid thistle]

that gives the amplitude from the midline

[flint stirrup]

mhm

[viscid thistle]

giving both the min and max

[flint stirrup]

exactly

what about the 8?

[viscid thistle]

hmm..

does that effect the horizontal shifts?

[flint stirrup]

so we know the amplitude is multiplied by 3

[viscid thistle]

yea

[flint stirrup]

this is the problem you have

[viscid thistle]

does it make the periods shorter?

[flint stirrup]

mhm!

in your equation, what is B?

[viscid thistle]

1

[flint stirrup]

not quite

[viscid thistle]

or 8...

[flint stirrup]

8

: )

so we know the amplitude is 3

we can find the period with this

[viscid thistle]

oh man ive never seen that before

ok

[flint stirrup]

what would your period be?

[viscid thistle]

1pi/4

[flint stirrup]

mhm

now

what is D in your equation?

[viscid thistle]

5

which raises my starting point?

which was at 0

[flint stirrup]

so we have

middle = 5
period = pi/4
amplitude = 3

exactly

ok so if the middle of the graph is 5

and the amplitude is 3

what are the min and max?

[viscid thistle]

max=8 and min=2

[flint stirrup]

right on!

[viscid thistle]

that made me sweat lol

[flint stirrup]

now lets revise what know so far,

middle = 5
period = pi/4
amplitude = 3
min = 2, max = 8

we can almost draw the graph now

we just know where to start

you ever see this "-C/B"

[viscid thistle]

yea

[flint stirrup]

cool

so where would we start? (x,y)

you already know the y

try to find the x first

[viscid thistle]

(-1/2, 5)

?

[flint stirrup]

the y is right

this is how you find the x

what are C and B in your equation

remember that you have 4(x+8) in your equatoin

you might have to multiply that out

[viscid thistle]

oh shoot

ok

(-3, 5)

[flint stirrup]

close

so we multiply it out, and we get

4x + 32

so now, what is -C/B?

[viscid thistle]

oh snap

so -4

[flint stirrup]

mhm

[viscid thistle]

(-4,5)

is where i would start!

[flint stirrup]

nice 😄

!!!

now

middle = 5
period = pi/4
amplitude = 3
min = 2, max = 8
start = (-4,5)

lets start at the starting point

[viscid thistle]

the directions want two periods

[flint stirrup]

okay

so you can start with a point at the starting point

[viscid thistle]

would i plot the point on a regular graph?

[flint stirrup]

mhm

[viscid thistle]

ive been seeing different numbrs on the x axis

with radians?

[flint stirrup]

are you hand drawing this graph?

[viscid thistle]

yea

[flint stirrup]

ok so

draw the horizontal y axis

are first step is how should we label it (what is our scale)

[viscid thistle]

k

[flint stirrup]

you should start with the middle being your starting point

so that x of your starting point is -4

uh ill draw it

[flint stirrup]

o

nice

now

[viscid thistle]

using my phone and comp for this lol

[flint stirrup]

you should draw another point that is pi/4 away in both directions

since that is our period

the pattern would kinda restart and end up pi/4 to the left and right

[viscid thistle]

how do i know a distance of pi/4

on a normal graph

[flint stirrup]

you can actually graph with pi as the numbers

[viscid thistle]

0 being the origin?

oh dam

[flint stirrup]

mhm

i havent drawn the vertical axis

[viscid thistle]

i see

[flint stirrup]

we can make the vertical axis normal numbers since we arent dealing with pi as our y values

we know that pi = 3.14

so pi/2 would be around 1.5

so if we were looking to graph our starting point (-4,5)

the x value would be between -pi and -3pi/2

since -4 is between -3.14 and -4.6

[viscid thistle]

ok

[flint stirrup]

p

so there is our starting point

make sense so far?

[viscid thistle]

yea

[flint stirrup]

cool

now

[viscid thistle]

im following it

[flint stirrup]

we know that every period, the graph kinda restarts

it does the whole

and come back to the middle

[viscid thistle]

thats 180

the middle

then the 2pi is 360

[flint stirrup]

so we know the period restarts every pi/4

so we can draw a point at the middle pi/4 away, like this

shoot

i think the x scale is too big

[viscid thistle]

that makes sense

but i can just multiply each x number to 8 right?

so like pi/4* 1/8

[flint stirrup]

ok

[viscid thistle]

idk what im saying

ha

[flint stirrup]

new graph with better scale

-4 is pretty close to -5pi/4

and then we go 5 up, to find our starting point

[viscid thistle]

mmk

[flint stirrup]

oh

ok so

[viscid thistle]

o.o

[flint stirrup]

i accidentally went

autopilot and drew it

but thats okay, we can still see what we did

so we started at (-4,5)

and drew another point at that same y-value but pi/4 away

and in the middle between those two point, the graph would also cross the y-value

actually, do you have any questions in general, or does it not make sense?

[viscid thistle]

yea im getting it

i have to move the second point pi/4 away

sorry

[flint stirrup]

thats for a normal sine

because the period of a sine function is 2pi

[viscid thistle]

yea this was transformed

[flint stirrup]

oh

yea

so you go pi/8 away to find where it first crosses

then go another pi/8 away to find where it ends

which makes up the whole pi/4 period

and then you do it again for the 2 periods you need

[viscid thistle]

awesomesauce

my hw should look a lot better now

lol

now that i know how the periods work

and the transsformations

i wish i can give you a medal lol

[flint stirrup]

yo

😄

[viscid thistle]

Hello

What phase shift does

y = -1/5cos((x/3)-(pi/6)) +4

have

?

[stuck lark]

if you have a graph of a trig function, think of it as a "horizontal" shift

[viscid thistle]

I put down

I put down right Pi/6 units

Yeah

But the thing is I got that wrong on the test

[stuck lark]

could you send a picture of the question from the test?

[viscid thistle]

Sure

[stuck lark]

,rotate

[stuck lark]

the directions got cut off

[viscid thistle]

the rest says

including the change in period.

[scenic meteor]

evaulate e^ln(5x+6)

[stuck lark]

they're asking for ALL transformations with respect to y=cos(x) i assume

[scenic meteor]

and evaulate ln(6th root of e^x)

thank you!

[viscid thistle]

Yes

They are

[stuck lark]

can you list them off?

[viscid thistle]

Amplitude

Period

Reflection

Vertical Shift

Horitzamtal Shift

I got both period and phase shift wrong

But I know why I got period wrong

Just not for phase shift

[stuck lark]

@viscid thistle actually the shift isn't +pi/6

for example, for some function f, the graph of f(x-2) represents a horizontal shift of the graph of f(x) to the right 2 units

[viscid thistle]

so like

what would it be in this case then?

[stuck lark]

you want to rewrite the function in a way that you can see a term where a constant is being added or subtracted from x

[viscid thistle]

ah okay

i think i got it now

thanks

[stuck lark]

no problem

[tropic crown]

not sure where that negative came from

i think im missing something fundamental here

[uncut mulch]

they did something dodgy

there shouldve been a - sign in the line above

for x<0, x= -|x| = - sqrt(x^2)

[tropic crown]

@uncut mulch

if u dont mind, can u rewrite that in layman terms?

i dont really follow the rules here

that youre trying to imply

[uncut mulch]

do you understand the properties of absolute values?

[tropic crown]

hmm, somewhat, but ill review it rn

thanks for giving me a hint

@uncut mulch okay, i reviewed absolute values properties, but unfortunately, i was unable to understand why this applies to my limit question

[uncut mulch]

did you understand

for x<0, x= -|x| = - sqrt(x^2)
@tropic crown

$\lim_{x\to -\infty} \frac{2x-1}{\sqrt{3x^2+x+1}} =\lim_{x\to -\infty} \frac{2-1/x}{-\frac{1}{\sqrt{x^2}}\cdot \sqrt{3x^2+x+1}}$

[obsidian monolith]

ramonov:

[uncut mulch]

$= \lim_{x\to -\infty} \frac{2-1/x}{-\sqrt{3x^2/x^2+x/x^2+1/x^2}} $

[obsidian monolith]

ramonov:

[uncut mulch]

the initial indeterminate form also had the signs in the wrong places (-inf/inf)

[crimson thorn]

yo i have a question
how do you find trig functions without a calculator
for example sin(5pi/3)
like ik u have to sketch it but i don't understand how u sketch it when it's given in radians

[vernal rapids]

Get a coordinate system

Mark 0, pi/6, pi/2, and pi on the x axis
Reflect pi/6 over pi/2 to get 5pi/6

After you marked those, shift everything by pi to the right to get
Pi, 7pi/6, 3pi/2, 11pi/6, 2pi

Now you have (almost) all important points

But with these you can draw a sketch

Sin(0)=0, so you mark point (0,0)
Sin(pi/6)=1/2, so you mark (pi/6,1/2)
Sin(pi/2)=1
Sin(5pi/6=1/2
Sin(pi)=0
Sin(7pi/6)=-1/2
Sin(3pi/2)=-1
Sin(11pi/6)=-1/2
Sin(2pi)=0

Not you got 9 points that's on the graph sin(x)
Connect them nicely and you got a sketch of sin(x)

I will make a drawing in a minute to show what I mean

[uncut mulch]

do you know the relation between radians and degrees?

[river coral]

anyone knows how to find the domain for piecewise functions? Im not sure how to (i dont think we gotten to that part for the lecture, idk) I did the graphing part, check if its right

[willow bear]

the domain is the set of all allowable x values

but also your sketch is off

and not actually representative of the function whose formula is given up there

[river coral]

the only part i get confused which is the domain so how would you interpret it based on the graph where the lines showed up like this?

or i did the graphing wrong

all i know is that the domain can be infinity (+ and -)

:(

i did the solving using the help video provided and based it off of it

so i did it wrong

f

is it off?

[willow bear]

yes it is still off

[river coral]

hold up i need to show you what i did

Did I do the solving correct?

Ignore the writing

[uncut mulch]

from your working out
at x=-1, f(x)= 0
at x=-2, f(x) = -1
but that isn't represented on your plot

[river coral]

so did i basically did the solving part correct but plotted it wrong?

[uncut mulch]

also did the program cut off those lines at x=-5 and x=4?

[river coral]

ah no i can make it long as it allows me to like to the edge of the graph

[uncut mulch]

is the function defined past those points?
should the lines continue?

[river coral]

it should? because there isnt a limit set on how long it should go? (except for the starting points)

but i do know that certain expectations it stops....

like it has another point

but thats only if it curves

so because its linear it should hit the edge of the graph?

[gusty igloo]

I had a question but by the time that this discord let me talk I figured it out lol

Thanks for the help bois 👌

[native timber]

the classic rubber duck method

[low lily]

anyone have any idea how to begin to solve this: f(t) = 2t+8 / t - A and f(-1) = -2

[uncut mulch]

do you understand what f(-1) means?

[low lily]

honestly im not sure what it means

[willow bear]

do you know what a function is

[low lily]

i do

[willow bear]

and what it means to evaluate a function at a point

[low lily]

well i understand what it means when it says f(-1) = -2

[willow bear]

so then what do you not understand

actually why don't you post the exact statement of the problem

and tell me what part of that you're having trouble with

[low lily]

oh im sorry. It wants me to solve for A

[willow bear]

the exact statement of the question

in its entirety.

do you have a screenshot?

[low lily]

one sec

[willow bear]

okay

well

what happens when you plug in t=-1 into f(t)?

[low lily]

i could solve for A

[willow bear]

let me be more clear.

what do you get upon plugging in t=-1 into f(t)?

[low lily]

2(-1)+8 / -1 -A

?

[willow bear]

parentheses.

you didn't mean $2(-1) + \frac{8}{-1} - A$, did you?

[obsidian monolith]

Ann:

[low lily]

no no

[willow bear]

then you really should use some parentheses

(2(-1)+8)/(-1-A)

this is what you meant

[low lily]

yeah

[willow bear]

well

now set that equal to -2 and solve for A!

[low lily]

ahh ok

thank you so much @willow bear

thanks again @willow bear I really appreciate the help

[summer sierra]

In the year 1985, a house was valued at $114,000. By the year 2005, the value had appreciated exponentially to $140,000. What was the annual growth rate between 1985 and 2005? (Round your answer to two decimal places.) Can someone walk me through this process?

[willow bear]

let r be the annual growth rate

the house appreciates exponentially, so each year its value gets multiplied by (1+r)

[summer sierra]

i have no idea what im doing wrong i got 0.01 as the annual growth rate

but idek

[willow bear]

can you show your work?

[summer sierra]

yes

[willow bear]

oof your work is a bit of a mess

[summer sierra]

sorry

[willow bear]

i can see you have arrived at (1+r)^20 = 140/114 though

[summer sierra]

yes but when i take the 20 root

[willow bear]

,calc (140/114)^(1/20)

[obsidian monolith]

Result:

1.0103251388592

[summer sierra]

then i subtract them from 1

and then it just gives me 0.01

[willow bear]

so r = 0.0103 = 1.03%

[summer sierra]

omg thank you

[hoary valley]

Hey guys, Are these the same ?

[blazing parrot]

yes

[leaden stratus]

Is the procediment correct or do I need to do it in another way?

EDIT: I no longer need it.

[heady jewel]

here

[gusty seal]

Hi, I have a question regarding sequences. I know the 5th term of this sequence and the sum of the first three terms. I need to find the common difference and price that the first term is - 6.

Thanks for any help.

Prove instead if price*

[heady jewel]

can you post the exact problem

with the values

[gusty seal]

Ok I'll try to find which page of the textbook it is

My friend sent me a picture but it's really blurry.

[heady jewel]

ok

[gusty seal]

Found it

Sorry back now my internet had issues

Phew the image finally seems to be sending now

[cold oyster]

wasnt trolling but ok

[gusty seal]

On this page it's question 8 and 9, would really appreciate any help.

Oh

Hmmm

@helpers

[uncut mulch]

what have you tried for 8

[gusty seal]

I'm not really sure what to do

I missed the class where we were taught this and am not very good with sequences and series in general

[uncut mulch]

what do you know about arithmetic sequences/series?

[gusty seal]

Not sure about which topics I don't get, I just don't really know how to take on this question.

[uncut mulch]

do you know the expression for the nth term?

[gusty seal]

Not quite sure what you're talking about

[uncut mulch]

or the relationship between consecutive terms

[gusty seal]

Like x+kn?

[uncut mulch]

what's x k and n?

[gusty seal]

The base number would be x, n would be which term it was, and k would be the number multiplied by n

And I know how to use the sum of terms and nth term functions

[uncut mulch]

close. that's not quite the expression for the nth term.

[gusty seal]

Hmm

Is it n+1 or something

[uncut mulch]

wrong direction

whats your first term?

[gusty seal]

I'm honestly not sure

[uncut mulch]

The base number would be x

[gusty seal]

About the standard formula

X?

Is it different for linear and exponential sequences?

k*n + x?

[uncut mulch]

you used x and k but usually a and d are used for the first term and difference respectively

k*n + x would give the (n+1)th term

[gusty seal]

Oh thanks

That sounds familiar

[uncut mulch]

the first term is a
what would be the 2nd term?

[gusty seal]

a + k

[uncut mulch]

yeh, preferably use d from now on

3rd term?

[gusty seal]

Ok

a + 2*d

[uncut mulch]

how many d would you need to add to a to get to the nth term?

[gusty seal]

(n-1)*d

[uncut mulch]

yep

[gusty seal]

:)

[uncut mulch]

so using that info, what's the expression for the nth term?
T_n =

[gusty seal]

(n-1)*d + a

[uncut mulch]

ok good.

are you able to get an equation using the information the 5th term is 14

[gusty seal]

(5-1)*d +a = 14

[uncut mulch]

can that be simplified?

[gusty seal]

Yes, to 4*d + a = 14

[uncut mulch]

yep and that will be your first equation (which you will need later)

do you know the formula to calculate the sum of n terms?

[gusty seal]

I know how to use it but can't write it off by heart.

[uncut mulch]

or describe it in words

[gusty seal]

Yeah I know it uses the initial term and the common difference

It's in our formula sheet

[uncut mulch]

try your best to transcribe it

[gusty seal]

It's (n (a * a subscript n)) /2

[proven dove]

how necessary is precalc for calc?

[undone pawn]

an easy form of the sum is (n/2)(a+l)

[gusty seal]

But I'm not sure what a subscript n is

(a+l)?

[uncut mulch]

is that * meant to be a +?

[gusty seal]

Oh yeah

Oops

[undone pawn]

first term + last term

[gusty seal]

Oh ok

[uncut mulch]

a_n is the nth term

[undone pawn]

basically the sane thing lol

[uncut mulch]

and a(_0) is the first term

[gusty seal]

Oh ok makes sense

[undone pawn]

yeah

[uncut mulch]

here your expression for the nth term was (n-1)*d + a

[gusty seal]

So we can set up simultaneous equations right

[uncut mulch]

$S_n = \frac{n}{2} (a + (n-1)*d + a) = \frac{n}{2} (2a + (n-1)*d ) $

[obsidian monolith]

ramonov:

[gusty seal]

Ok I get this now

So we can also get (3/2) (2(-6) + (3-1) * d)

Which simplifies to 1.5(-12 + 2d)

=-3

3d-18=3

And 4*d + a = 14

[uncut mulch]

was this for part a)?

[gusty seal]

Anyways thanks for the help I'm going to try to work the rest out now

Oops I made a mistake

Oh you're back

[uncut mulch]

you aren't allowed to substitute a=-6 yet

[gusty seal]

Yes, but I realized I made a mistake there

Yeah

I misread

So it should be (3/2) (2a + (3-1) * d)

[uncut mulch]

is equal to what?

[gusty seal]

Is equal to _3

= - 3

[uncut mulch]

yep

[gusty seal]

So (1.5 * (2a + 2) * d) = - 3

And 4d + a = 14

[uncut mulch]

top equation is a bit inaccurate

misplaced parnetheses

[gusty seal]

Oh, is it (1.5 * 2a + 2 * d) = - 3

Because they would be around the 2

[uncut mulch]

that's still inaccurate because the 3/2 or 1.5 applies to the whole of (2a + 2d)

[gusty seal]

Oh yeah that should be outside the brackets

Oops

So 1.5 * (2a + 2 * d) = - 3?

[uncut mulch]

don't need that * between the 2 and the d
do you think you can simplify the equation?

[gusty seal]

Can be simplified to1 (3a + 3d) = - 3?

Then a+d = - 1

[uncut mulch]

yep.
can you handle the rest?

[gusty seal]

I think I can yeah

Thanks for the help

[uncut mulch]

np

[gusty seal]

The other questions should be similar so I'll be good from here :)

[gusty seal]

Hi, is anyone able to help with the second part of that question? I'm a bit unsure about how to solve it.

It's question 8b

[uncut mulch]

what was your value for the difference?

[uncut mulch]

a = -6
d = ?

[gusty seal]

I worked out d = 5

[uncut mulch]

what would be expression for the nth term now that you know a and d?

[gusty seal]

5(n-1)-6

[uncut mulch]

how would you find values of n where that is greater than 282?

[gusty seal]

I kind of don't get the question

Is it asking me for the smallest number outputted from that equation while the n put in is greater than 282?

[uncut mulch]

its asking for the smallest n where T_n > 282

[gusty seal]

Ok

I feelike I'm making a mistake because the higher I make T_n, the lower n becomes

So when T_n = 283, the nth term would be 5(583-1)-6, right?

[uncut mulch]

ah thats the wrong approach.

[gusty seal]

Oh, how should it be done then?

[uncut mulch]

to find n when T_n > 282, you would need to solve that inequality.
(and you also have the expression for T_n)

[gusty seal]

So just to be clear T_n is the n inside the equation?

I haven't been taught this notation

[uncut mulch]

T_n is the nth term

[gusty seal]

Ok

So that's what the answer to the equation is?

[uncut mulch]

not an equation but an inequality
you want to solve T_n > 282
and you also know that T_n = 5(n-1)-6

[gusty seal]

Ok, makes sense now.

I think I got n and T_n mixed up

So is the next step to do 5(n-1)-6>282?

[uncut mulch]

yes

[gusty seal]

I simplified it to n>58.6

So does thad mean T_n is lowest when n=58.6?

[uncut mulch]

not quite.

[gusty seal]

Hmm, I don't think that's correct because I just end up with 282 again

Yeah

[uncut mulch]

what's a requirement of n in sequences?

[gusty seal]

Does it have to be a whole number?

Or a natural number?

Because I'm pretty sure they can't be negative

[uncut mulch]

yeh.

and whats the smallest number >58.6 that fits that requirement

[gusty seal]

That would be 58

[uncut mulch]

greater than 58.6

[gusty seal]

Oh oops

59 then

[uncut mulch]

if you plug in n=58 you'll get a number smaller than 282

[gusty seal]

Ok thanks

I get it now

[viral imp]

so I'm looking at this:

https://www.wolframalpha.com/input/?i=log_3(x)%3D-4%2F9

not sure how the solution gets from

,, \frac{log(x)}{log(3)}=-\frac{4}{9}

[obsidian monolith]

Umma.Gumma:

[viral imp]

to

,, \frac{1}{3^\frac{4}{9}}

[obsidian monolith]

Umma.Gumma:

[viral imp]

I basically follow it until this step:

,, x = 3(-\frac{4}{9})

[obsidian monolith]

Umma.Gumma:

[summer sierra]

An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let x = length of the side of the base

can someone help with this

[modern nymph]

You meant root?

[summer sierra]

no thats the exact wording of the problem

one of the sides is open due to it being an open box

[modern nymph]

Just say cubic box is a^3 you know x=a?

[merry sphinx]

The box is not necessarily cubic

[modern nymph]

Hmm

[merry sphinx]

Only a square base, says nothing about height

[summer sierra]

but its in cubic inches

so idk what to do

[merry sphinx]

Volume is in cubic inches

FairyWT thought the width/height/length needed to be the same

[summer sierra]

oh

[merry sphinx]

Does it specify if the side lengths are integers?

[summer sierra]

it says construct a rational function to help solve the problem

[quick lintel]

A rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let x = length of the side of the base. Construct a rational function to help solve this problem.

I require assistance

[viscid thistle]

Question

How do u calculate lim x-> neg. Inf of

[cobalt totem]

@viscid thistle Hey! I would graph it and look at the end behavior of the graph

or

=tex \sqrt{x^2 + 5} - x

oh uh

didn't work but

• (-inf) is pos inf

so

[stuck lark]

$\sqrt{x^2+5}-x$

[obsidian monolith]

RokettoJanpu:

[cobalt totem]

eyah

$\sqrt{(\inf)^2 + 5} - (-\inf)$

[obsidian monolith]

silencesilence:

[cobalt totem]

is

just looking at what happens to infinity

well, infinity times infinity is infinity

if you add five to infinity it still goes to infinity

sqrt of infinity is the same

minus minus a number is plus the number

so you have infinity plus infinity

[flint river]

what if..

$\lim_{x\to +\infty}$

[obsidian monolith]

ike98:

[flint river]

do you factor sqrt(x^2) or do "conjugate" first?

I'm asking cause it ain't intuitive to me 😦

[stuck lark]

conjugate is the way to go

[cobalt totem]

do the conjugate?

Can you show me what u mean

[stuck lark]

since you mentioned conjugate yourself, do you know what it means in this context?

[flint river]

who, me?

[stuck lark]

thought you two were the same person, sorry

[flint river]

😄

[cobalt totem]

can you guys explain that to me

[stuck lark]

silence, as suggested, you should evaluate the limit as x approaches inf. naively plugging in x = inf gets you nowhere

[bright comet]

can some help me with my precal question?

[stuck lark]

it pretty much leads to inf - inf, so some algebraic manipulation of the limit is required

[cobalt totem]

yea, that's true, buit if you can see it doesn't lead to inf - inf

[stuck lark]

$\lim_{x \to \infty} \sqrt{x^2+5}-x$

[obsidian monolith]

RokettoJanpu:

[cobalt totem]

yea, so how would you do that one?

yea

[obsidian monolith]

ike98:

[stuck lark]

as suggested before, multiply the expression by the conjugate

[cobalt totem]

ahhh

OH

[stuck lark]

but to avoid actually changing the limit, divide by the conjugate as well

[flint river]

I had an exam today with almost the similar problem and failed to solve it 😦

[cobalt totem]

so (sqrt(x^2 +5) + x)/(sqrt(x^2+5) + x)

is what you multiply it by

[stuck lark]

right

[cobalt totem]

so then

how do you get rid of the denominator

[stuck lark]

what do you have so far?

[cobalt totem]

well what 'd you get is

((x^2+5) - x^2)/(sqrt(x^2+5) + x)

[stuck lark]

simplify

[cobalt totem]

well

5/(sqrt(x^2+5) + x)

[stuck lark]

take lim x -> inf

[cobalt totem]

OH

0

[stuck lark]

👍🏽

[flint river]

check this out

[cobalt totem]

I see i see

but i'd probably just graph it instead

[flint river]

let me write how I had it today:

[stuck lark]

if i asked you to find the horizontal asymptotes of the function without a graph, this is the way to do it

[cobalt totem]

ye

[flint river]

$lim_{x\to+\infty} \sqrt{x^2-3x+6}-x$

[obsidian monolith]

ike98:

[cobalt totem]

let's see

let me try that

[stuck lark]

\lim tbh

[flint river]

indeed i'm getting errors everywhere duh 😅

[cobalt totem]

oh this one is just zero too

or wait

[flint river]

haha I thought the same

[cobalt totem]

you get -(inf/inf) which is undefined still

[flint river]

I was stuck and L'Hopitaled that sh** out of frustration

and felt guilty of course lmao

[cobalt totem]

how are you supposed to do this

[flint river]

what expression you got?

[cobalt totem]

(-3(x-2))/(sqrt(x^2 - 3x + 6) + x)

which gives you actually inf - inf again

so

wait

[stuck lark]

divide top and bottom by x raised to the highest degree

[cobalt totem]

if you complete the square on the bottom you get rid of inf - inf

[flint river]

how so?

[cobalt totem]

x^2 - 3x + 6 complete the square is (x-3/2)^2 + 15/4

so now it's just inf on the bottom

but it's still inf/inf

-inf/inf

[flint river]

😄

[stuck lark]

what's the highest degree on the top?

[cobalt totem]

1

[stuck lark]

bottom?

[cobalt totem]

2

[stuck lark]

but it's wrapped inside sqrt

[cobalt totem]

so 3/2

or

uh

oops

not that

haha

1/2

[stuck lark]

no

[cobalt totem]

what is it

1

?

[stuck lark]

[cobalt totem]

because

+x

ahhh

I forgot about that

haha

[stuck lark]

divide top and bottom by x^1 aka x

[cobalt totem]

why?

oh wait

you still get inf / inf

[stuck lark]

what's the top?

[cobalt totem]

-3 + 6/x

[stuck lark]

bottom?

[cobalt totem]

sqrt((x-3/2)^2 + 15/4)/x + 1

now you have inf/inf still

[stuck lark]

know how to move 1/x into the sqrt?

[cobalt totem]

sqrt((1/x)^2)

[stuck lark]

keep going