precalculus

[sudden bay]

evaluate f(-2)+f(0)-f(-3)

2x - 1, x < -2

x^2 + 7x, x = -2

|9x-5|, x>-2

what do i plug in first

[sudden bay]

i tried screenshotting it but it refuses to workk

[elder charm]

What's your question? Like, what do you have trouble with here?

[strange arch]

I think u don’t know much about piecewise functions

Just revise what they are

And u will be able to solve this

@sudden bay

[sudden bay]

how>

im confused because you are right

i dont know much

[uncut mulch]

the conditions on the right indicate when you should be using each piece

[sudden bay]

so i should only plug in the numbers given above when the stuff on the right is true

[uncut mulch]

ig

[sudden bay]

ok i did that

but it said the answer was 2

and i dont know how

[uncut mulch]

show your work

[sudden bay]

well my teacher provides the answer and he says its 2.

I put -3 in the first spot and replaced x with -3 so 2 * -3 -1
then -2^2+7x
then 9*0-5

and I added them but they do not = 2

well i added the -2 and 0 equations and subtracted the -3

[uncut mulch]

then -2^2+7x

could you make that more clear

firstly the value of x: "-2" is the thing being squared
i.e. (-2)^2
and why's there a still x in there

[sudden bay]

ya that was a mistake

i meant to put 92

-2*

the answer came out to be something like -4 though

[uncut mulch]

()()()

$-2^2$ is the negative of 2 squared \
$(-2)^2$ is the square of negative 2

[obsidian monolith]

ramonov:

[uncut mulch]

then 9*0-5
you ignored the absolute values

fix the mistakes and tell me the final values you get for each

[sudden bay]

I dont understand how

[uncut mulch]

when evaluating f(-3), since -3 < -2
you should be using 2x-1 right?

@sudden bay

[sudden bay]

Yes

[uncut mulch]

sub x=-3 to get your f(-3)

i.e $f(-3) = 2\cdot(-3) - 1 = , ?$

[obsidian monolith]

ramonov:

[sudden bay]

-7

[uncut mulch]

now similarly, for f(-2) you'd use the middle piece

$f(-2) = (-2)^2 + 7\cdot(-2) = , ?$

[obsidian monolith]

ramonov:

[sudden bay]

-10

[uncut mulch]

$f(0) = |9\cdot0 - 5| = , ?$

[obsidian monolith]

ramonov:

[sudden bay]

5

[uncut mulch]

and then plug those values into the original expression you're asked to evaluate

[sudden bay]

-7-10+5?

[uncut mulch]

no

note that in the question its **-**f(-3)

i.e. -(-7) = (+)7

[sudden bay]

-10+5+7?

I understand now

thank you

[fiery wren]

hi

can someone check me

im either doubting myself or overthinking lol

[blissful kayak]

Ooh these are fun!

Okay so

It is correct, but you can actually go one more step further...

You can split apart the 2x

But something tells me they don't care about that...

Yea just go for it

Click the one you already have highlighted for the first question

And for the second question your answer is right. It is inaccurate

[fiery wren]

phew okay

first time doing logarithms

and im already doubting myself

[blissful kayak]

They're not easy

[fiery wren]

so the first one

isnt complete right?

[blissful kayak]

Right

[fiery wren]

im looking at my notes

ok

alright lmao sorry

[blissful kayak]

All good

[fiery wren]

ive been overthinking about that one problem

for fifteen mins

oh and i have another question right

if im dividing the logs

ln A / ln B

is it equal to saying

ln(A/B)?

[blissful kayak]

Not quite

It's equal to log_B(A)

ln(A/B)=ln(A)-ln(B)

[fiery wren]

what about ln A + ln B?

[blissful kayak]

ln(AB)

[fiery wren]

ln(AB)?

or

ok

[blissful kayak]

Yuh

[fiery wren]

phew

[blissful kayak]

Honestly I find it pretty unintuitive that schools have you learning the properties of exponents and logarithms back to back

[fiery wren]

it's confusing

lemme change my name before i get in trouble

[blissful kayak]

Because you know multiplying the inputs for logarithms is the same as adding the outputs, and inversely, adding the inputs for exponents results in multiplying the outputs

[fiery wren]

so ln A / ln B

like you said is essentially

lnA-lnB?

just clarifying to update my notes lol

[blissful kayak]

ln(A/B)=ln(A)-ln(B)

[fiery wren]

got it

thank you

[blissful kayak]

Ye!

[terse ravine]

This doesn't seem right to estimate the value of f'(-1)

f'(-1) does not equal 2

[alpine scaffold]

@terse ravine are you approximating the value of the derivative of the graphed function at x = -1?

Just want to be clear

[terse ravine]

Using the graph of f(x) to estimate the value of f'(-1).

[alpine scaffold]

So which graph is in the picture?

f(x) or f'(x)

oh ok

thanks for the edit

So uh first thing you should do is approximate it around x = -1

By the picture it looks like youre approximating f'(1.5) ish

Try sampling f(-2) and f(0) and doing the same process

[terse ravine]

I got it right by accident and forgot how I did it.

[alpine scaffold]

lmao

Happens

[terse ravine]

Let's try another example.

[alpine scaffold]

Sure

Ok so pick points around x = -1 on the graph

Ones you can approximate accurately

What would you suggest?

[terse ravine]

-1 on the x axis and

the second part throws me off

[alpine scaffold]

hmm no

You want to be able to pick two points so that when you draw a line connecting them, the slope of that line is roughly the same as the slope of the graph at x = -1

Does that make sense?

In general, picking points that are one bigger and one smaller than the point you're approximating will work, unless the graph is especially bumpy

[terse ravine]

It does make sense, but why can't I see where to draw the line.

[alpine scaffold]

Try drawing a line between x = -2 and x = 0

haha no sorry

I mean f(-2) and f(0)

my bad

You want to approximate the slope based off points on f

not points on the x-y plane

if that makes sense

Thats f(-1) to f(0)

do f(-2) to f(0)

Remember you want to approximate the value at f'(-1) so you want to approximate the average rate of change around f(-1)

Here let me help

So the yellow line is +-1 around x = -1

the blue line is +- .5 around x = -1

As you get closer to points around x=-1 you'll notice that it approximates the slope much more accurately

Since you can easily approximate the value of f(-1) and f(0) I would choose those two.

[alpine scaffold]

beautiful

so now just compute

[terse ravine]

I think I did this right

[alpine scaffold]

looks about right yes

err

yeah

i think thats right

[strong fossil]

can someone explain this to me? pls ping me and thank you so much :'))

[lethal oracle]

Do you know how to find plots @strong fossil

[viscid thistle]

You can't plot the quadratic

I.mean it has no real roots

[strong fossil]

yeah so i thought it was just rad 2

but it's not :')

[viscid thistle]

it's can be done easily just take the roots to be a b c

a=sqrt(2)

For the linear factor

b and c are roots of the quadratic

use Vieta's

b + c = -(-√3)/1 = √3

So a + b + c = √2 +√3

Which is almost 3.15

[strong fossil]

oh woot

tysm! @viscid thistle

[viscid thistle]

Np

[patent junco]

Having a bit of an issue with vectors and how to work this problem:
u=<-8,-20> and w=<-3,-1>, find 1/4u-6w

[smoky pagoda]

hmm as far as I know there isnt really a definition for a vector's "multiplicative inverse"

i just dont see how that would make sense since vector multiplication isnt a thing

maybe they just want you to take the reciprocal of each elements?

[willow bear]

1/4u can only really be resolved as (1/4)u

[smoky pagoda]

oh of course

i just assumed everything was under the fraction

then in that case @patent junco what have you already tried

[unkempt magnet]

if a function f:R->R is defined by f(x)=x² is it neither injective nor surjective?
because f(2)=f(-2) so it cant be injective and the image of f(x) is R+ which is not equal to the codomain so it cant be surjective either

[viscid thistle]

Was that a question?

You asked it and answered it yourself

[smoky pagoda]

i think they just want us to confirm that they're correct

which they are

[unkempt magnet]

oh okay thanks

[narrow peak]

ohaiyo

The equation of a curve is $y=x³+3x²-9x+k$, where $k$ is a constant. Find the possible set of values for which the $x$-axis is a tangent to the curve.

[obsidian monolith]

Hmm:

[smoky pagoda]

are you familiar with calculus

[narrow peak]

yes

actually this is the 2nd part the first part wanted me to find the range of values for which y is decreasing

and i did

[smoky pagoda]

aight

so

using derivatives

have you already found maxima and minima points?

well tbf you can extract that easily from part 1

[narrow peak]

oh hold up

no i havent

[willow bear]

bruh

[narrow peak]

ive got an idea

[smoky pagoda]

do it for the value k =0

[willow bear]

have you found the points where y' = 0

[narrow peak]

yes

[willow bear]

you don't need to assume k=0 even lol

y' will be independent of k

[smoky pagoda]

no but so you get a y coord

and then get the offsets required straight away

[willow bear]

you'll have two points at which the slope of the tangent is 0

the value of k will then come into play to align those points vertically

[smoky pagoda]

^

[narrow peak]

oh

also i did sort of well for my math exam :D thx for the help last time @willow bear

[willow bear]

you just need to not overthink it

yw

[blissful kayak]

Back to the basics

Is this a function?

[willow bear]

no, it's a graph.

[viscid thistle]

what do you think

[blissful kayak]

I say so

Well

Just by observing

[narrow peak]

lold

[willow bear]

this could be the graph of a function, though, if that's what you were trying to ask.

[blissful kayak]

Yes

[willow bear]

but in math you have to be extra careful with how you phrase your questions.

[blissful kayak]

:/

Tue

*true

[willow bear]

the instant you phrase it with even the slightest hole, someone will come along and give you an answer which fits in that hole

and is technically correct but not what you intended

[blissful kayak]

So I think the graph could be a function

[patent beacon]

@blissful kayak
Yes, it's a function as it passes a vertical line test

[blissful kayak]

Thanks :3

[unkempt magnet]

for a function f:(-inf,0] ->[0,inf) defined by f(x)=x²
i wouldve thought this function would be bijective and thus invertible but does its inverse even work when the codomain of the inverse would be non-positive

[burnt sonnet]

Have you plotted it?

You will get f^(-1)(x)=-sqrt(x) I belive with domain [0,inf) and co-domain (-inf,0] ?

[unkempt magnet]

yeah i was thinking should i put - infront of it

but idk it seems off

[burnt sonnet]

Well have you thought about an inverse as rotating the function around y=x axis?

[unkempt magnet]

yes

[burnt sonnet]

does it make sence then to get -sqrt(x) ?

[unkempt magnet]

but what about when you want to convert it back

[burnt sonnet]

ye

what is the issue?

[unkempt magnet]

it doesnt feel correct even though it makes sense

[burnt sonnet]

Hmm

I mean i get it from definition and thinking about rotating it around y=x axis

[unkempt magnet]

because youre manipulating the rule of the function so that it satisfies the codomain

[burnt sonnet]

ok how do you solve for the inverse function?

[unkempt magnet]

wdym

[burnt sonnet]

So when i try to find the inverse function i try to solve x(y)=finv(y)

so you start with y=x^2 right?

[unkempt magnet]

oh

yeah the thing is

if i did f(f^-1(x))

[burnt sonnet]

to find the inverse i take $x=\pm\sqrt(y))=f^{-1}(y)$

[unkempt magnet]

that would be (-x^0.5)² which is x

so that works right

ah

[obsidian monolith]

egocarpo:

[unkempt magnet]

so -√x it is theb?

for that codomain

[burnt sonnet]

and then you have to see that only - solution works

yes

good

Did this help clearify it?

[unkempt magnet]

yeah thx

[burnt sonnet]

awesome!

[umbral cloud]

While A<=>B is true...
Is it sufficient to say A=> B is true?
For example...
A-B = 0
Can I say => A=B?
Or is it necessary to say <=> A=B?

[hearty night]

No, it's not necessary.

[patent beacon]

A ⇔ B

Can be thought of as both of
A → B
B → A

[half star]

Depends on what you want to prove or conclude. Sometimes you might need to prove an "if and only if" kind of statement and then <=> might be useful. Otherwise the implication => is fine.

[patent beacon]

I'm not certain on what exactly your question is haha

[umbral cloud]

@half star give me an example in which i should use <=>.... And thx everyone

[lime cargo]

Do you know where is the illegal step? Im convinced its the one in the blue square but i dont know how to explain it

[uncut mulch]

$\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a,b \geq 0$ \
$\sqrt{x}\sqrt{x} = x$

[obsidian monolith]

ramonov:

[lime cargo]

Ok now i see it

Thanks

[patent beacon]

@umbral cloud
As an example of using ⇔:

A triangle with side lengths a,b,c (longest side c) is a right triangle iff a² + b² = c²

So note there's essentially two statements cooked into one.

• A right triangle obeys pythag
• Any triangle that obeys pythag is right

[umbral cloud]

So... Is it false to say... a² + b² = c²
=> ABC is right triangle?

And thank you very much for the response

@patent beacon

[patent beacon]

Nope! That's true!

The second statement there

Any triangle that obeys pythag is a right triangle

And any right triangle obeys pythag

Therefore
Obeys pythag ⇔ Right triangle

[umbral cloud]

Understood... Thx a lot

[patent beacon]

Np, feel free to ask if you have any other questions on it!

[river moth]

So say if I was doing a test, would someone be able to help me cheat through it?...

[glad pasture]

how do you do this

[fast geode]

here's a start:

[obsidian monolith]

longhaul:

[glad pasture]

thanks i got it

how do you simplify this i got to x^-14 = x^3n+6

not sure if i did that right

[quick mirage]

x^n / x^15 = x ^ (n-15)

x^(3(n+2)) = x^(3n+6)

Lets update the equation

x^(n-15) = x^(3n+6)

Now using our knowledge of logarithms, we know that log b (b) = 1, just apply the same rule here but with base x

log x (x^(n-15)) = log x (x^(3n+6))

(n-15) log x (x) = (3n + 6) log x (x)

^ You can just skip this if you just assume n-15 = 3n+6 because they have the same base.

n-15 = 3n+6

And the rest is pretty self explanatory @glad pasture

[glad pasture]

why dont you multiply the 2 by 3 ?

i pluged in -21/2 and the equations were equivalent

[quick mirage]

right its 6 whoops

should be good

[glad pasture]

could you help me with this i got to ln(x+1)(x^2) = ln(x)(2) but not sure what comes after

[quick mirage]

Prove right side = left side type of problem?

or solve for x?

[glad pasture]

solve for x

[quick mirage]

we know log a + log b = log ab

as u have shown

ln ( (x+1) * (x^2)) = ln (2x)

ln( x^3 + x^2 ) = ln(2x)

0 = ln( x^3 + x^2 ) - ln(2x)
0 = ln((x^3 + x^2) / 2x)

The only number that can make ln(n) = 0 is if n = 1

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

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

1/2 (x^2) + 1/2(x) - 1 = 0

Using quadratic formula, we can find the solution.

@glad pasture

The coolest thing about the result is that it is quite literally the golden ratio!

Which is always cool

[glad pasture]

dont you have to factor out the x^3

[quick mirage]

no, because it gets simplified into x^2 when divided by 2x

well (1/2)x^2 to be precise

[glad pasture]

did you divded both equations by ln(2x) ?

[quick mirage]

Nope, I just subtracted on both sides

log a - log b = log(a/b) after all

[glad pasture]

i see now

[quick mirage]

hope I helped ya out

[glad pasture]

on the key it says that x = -2 as a root but you cant get 3 roots from a quadratic

[quick mirage]

the resultant value if you plug in -2 is likely a complex number

which i did not consider

Are there two other answers though? @glad pasture

[glad pasture]

i mean i think i need to state somewhere they are extranous solutions

because 1 is the only solution

[quick mirage]

I dont think my math is wrong either

its really a matter of how you tackle the problem, I guess, which is kinda dumb if the key doesn't consider other solutions

[glad pasture]

ok well thanks for the help

[wind seal]

ALERT: This is part of a calc homework about derevatives of tower functions. In a video explaining how to do them the instructor glossed over some log properties without mentioning the one.

How does $x^{ln(x)}$ = e^{ln^2(x)}

[obsidian monolith]

Noby707:
Compile Error! Click the reaction for details. (You may edit your message)

[wind seal]

I tried playing around and I got $x^{ln(x)} = ln^2(x)$ But how do we raise e to that.

[obsidian monolith]

Noby707:

[viscid thistle]

You want the derivative of x^lnx right ?

[wind seal]

Yes, I am trying to set it up to $e^{ln(x)^2}$, from there I can take any $d/dx[a^x]$.

[obsidian monolith]

Noby707:

[viscid thistle]

Yea that's right

that's easy derivative of e^f(x)

[wind seal]

Yes but how do I go from $x^{ln(x)}$ to $e^{ln^2(x)}$ with log manipulations?

[obsidian monolith]

Noby707:

[wind seal]

Calc is so easy but I feel stupid when steps involve some stuff I forget.

[viscid thistle]

you can use definition of log or just put y=x^lnx and apply logarithm of both sides so you get lny = (lnx)^2 which means y = e^( (lnx)^2 )

[wind seal]

oh, Thank you!

I forget to manipulate both sides of the equation instead of one side 😅

[viscid thistle]

That gives you degree 3

Obviously it can't be degree 4 and 2

[shadow plaza]

so factors of (x+2),(x+1)
@umbral current (x-1)

[willow bear]

-2 has even multiplicity as a root

what's the smallest positive even integer? @umbral current

[shadow plaza]

double ping. btw what is multiplicity?

[willow bear]

the multiplicity of a root r of a polynomial P is the exponent of (x-r) in the factorization of P

alternatively, it's the smallest positive integer $m$ such that $\dv[m]{P}{x} , (r) \neq 0$

[obsidian monolith]

Ann:

[shadow plaza]

Ohh yeah. Ofc.

🤦‍♂️

[willow bear]

note that this definition can be stretched to say non-roots have a multiplicity of 0

yeah so

(x+2)^2

you have f(x) = a (x+2)^2 (x-1)

[willow bear]

negative doesn't mean has a minus sign in front

[timber holly]

Can anyone teach me how to solve this kind of question? And no... this is not a test but a homework

[viscid thistle]

Where exactly are you stuck?

[blissful kayak]

-2 has even multiplicity as a root
@willow bear Yes

[uncut mulch]

yes

[patent beacon]

That's my approximate tangent line. You want the slope of that @terse ravine

[copper breach]

Has anyone taken the BYU course for Precalculus or is currently doing it?

[terse ravine]

(7.5-4.5) / (4-3) = 3

[viscid thistle]

@terse ravine remember the definition of derivative:

d/dq [h(q)] = h'(q) = [h(q+a) - h(q)]/a such that a is a little number (the difference of q)

then i think you could use:
q1 = 1,6
q2 = 2

q2 - q1 = a = 0,4

h(q1+a) - h(q1) = h(q1+q2-q1) - h(q1) = h(q2) - h(q1) = h(2) - h(1,6) = 402 - 188 = 214

[h(q1+a)-h(q1)]/a = 214/0,4 = 535

you can also do it with q1 = 1,2 and q2 = 1,6, i guess

[prisma vine]

I wasn’t sure if I did this right

[terse ravine]

[prisma vine]

Yo can someone help

[rustic bluff]

i can help

If u add me i can help anytime for small amounts of venmo or good anime recs

except not venmo cause i just read not not allowed

so anime recs only lol

[prisma vine]

Nah I figured it out lol

[cobalt dirge]

Hello , im an 11th grade student in Australia currently studying calculus, just cant figure this question out. Any help is appreciated! Thanks! y=x+1 is a tangent to y=ax^2+bx at the point (1,2) find a and b

[willow bear]

do you know what a derivative is?

there are two different solution paths depending

[cobalt dirge]

ohh sorry @willow bear had to leave for a minute

but yes I know all about the basic calc stuff, just confused on how to solve for a and b

because my instinct is f'(1)= 2ax+b

but i couldnt get any further

[willow bear]

you're mixing things up

first off there's f(x) = ax^2 + bx

[viscid thistle]

hi, I’m new I just found this discord server on Pinterest

[willow bear]

and then when you differentiate it you get f'(x) = 2ax + b

and then once you plug in 1 you get f'(1) = 2a*1 + b @cobalt dirge

[viscid thistle]

I was wondering if anyone knows anything about guess&check method

I’m new to precal

just started 2 days ago

and I’m really stuck on it

[willow bear]

this channel in particular is a bit occupied at this moment specifically...

[viscid thistle]

my bad

[cobalt dirge]

lol

how does it work from there then, f'(1)= 2a +b

[willow bear]

well what is f'(1), given that the line y = x+1 is the tangent at x=1?

[cobalt dirge]

1

[willow bear]

great

[cobalt dirge]

ohh ok lol i was thinking about it wrong,

[willow bear]

and what is f(1)? both in terms of a and b, and its actual value.

[cobalt dirge]

just had to visualise it

you mean re-arranged

[willow bear]

i mean exactly what i said

on the one hand, f(1) = a + b by the formula for f
on the other hand, f(1) = 2 because of the tangent line

[cobalt dirge]

ok

[terse ravine]

Well thats wrong hmm

I don't know what to do.

[viscid thistle]

@terse ravine maybe (535+250)/2

(the average?)

[terse ravine]

The question is above.

h'(1.6) = estimate

That's why I put 535 and 250.

[viscid thistle]

@terse ravine yea and that is what i meant

using estimate: (535+250)/2

but i'm not sure

[terse ravine]

393

I'm going to sleep on it.

[viscid thistle]

yes

[terse ravine]

Already 9 hours of Math problems.

[viscid thistle]

i'm in this server since mmm

maybe 4 hours

or 6h

[copper breach]

can someone help me with this question

what i a conjugate axis?

can someone explain the process of solving this please

[void tinsel]

Hm trying to figure this out

https://gyazo.com/0e23385b4199f2dbed0d2412e188d35e

I know it would be Cot

[potent pine]

can someone please ping me and help me with this

i’m leaning towards A on #1 but i’m not 100%

[viscid thistle]

Hmmm

[wheat plaza]

Is that a test?

[potent pine]

review self test

our test is next class

i’m just confused about all the concave stuff bc my teacher he briefly explained it

[viscid thistle]

A is fine

[potent pine]

is 2 A aswell?

i’m kind confused on how it goes from fast to slow again l

like how that changes the graph

[proud raven]

@potent pine sorta sounds like a logistic curve to me

the number of people that have heard the rumor over time, i mean

[gilded brook]

Currently I got f(x)=3sec(2pi/5x)-1 how do I find the phase shift?

[fast marlin]

The phase shift is inside your trig function, it's whenever you have + or - a constant inside the trig function

In this case, the argument of the trig function is (2pi/5)x, so there is no phase shift

If it was sec(2pi/5x + 3), the phase shift would be the +3 bit

[gilded brook]

Ah

So I've done something wrong in my base equation, I'm trying to find the equation for the graph shown

[gilded brook]

Figured it out period was 4 not 5

[viscid thistle]

I got 457 would that be correct

[gilded brook]

What is this asking?

[old flame]

@gilded brook did you get an answer yet?

[gilded brook]

Nah

And I'm trying to figure out how to do it fully

[old flame]

Um let me see

f(x) ^ 2

1/f(x) is the same as f(x)^-1

[gilded brook]

I think its taking the inverse and plugging in

[old flame]

My brains a bit fuzzy

not sure I will be ab le to help

able

[gilded brook]

Like can sub left side into the f(x) if you make the left side =f(x)

[old flame]

I'm just procrastinating from my own probs

[gilded brook]

dang

[old flame]

That's what I would try

Do this

hang on... f(x)^2 , I think is the same as saying, ((f(x)^-1)^2) + sin^2(x)

so, it's (f(x) to the power of -1)^2) + sin^2(x) ... then evaluate

[night fiber]

The function is even, and when saying tan=sin/cos I found (cos²+sin²)/cos =1/f

I might be wrong

[gilded brook]

I need to simplify the left into a single trig function I think

[old flame]

oh ok

[gilded brook]

Then I can sub into f(x)

And work it out

[old flame]

on it's sec(x)

oh it's sec(x)

[night fiber]

Where I am wrong if I say $cos(-x)+tan(-x)sin(-x)=cos(x)+tan(x)sin(x)=\frac{cos^2(x)+sin^2(x)}{cos(x)}=\frac{1}{cos(x)}$?

[obsidian monolith]

Svet L'octogone:

[old flame]

cos(-x) + tan(-x)sin(-x) evaluates to sec(x)

[night fiber]

Nice

[old flame]

🙂

[night fiber]

So the weird thing becomes 1

[gilded brook]

Is cos(-x) the same as -cos(x)?

[night fiber]

No

Look on a graph

You will see cos(x)=cos(-x) (cos is an even function) and sin(-x)=-sin(x)

You can see it through the unit circle too

[gilded brook]

How do you see on unit circle?>

And whats the graph called?

[old flame]

Examine x = 1 for all cos(x), sin(x) etc and compare the values as they change via altering the sign of x

-x , +x

[night fiber]

Rip PNG

Well... when the point t travels the unit circle, the x component is cos(t) and the y component is sin(t)

[gilded brook]

But where do you see those three with those +/- = sec(x)?

[night fiber]

You see that t and - t are symmetric with respect to the x axis, then cos(t)=cos(-t)

And with the same reasoning, sin(t)=-sin(-t)

[gilded brook]

Yeah

I see that

[night fiber]

tan is defined as sin/cos

So since sin is odd and cos is even, sin/cos is odd then tan(-x)=tan(x)

[gilded brook]

But secx=1/cosx

Right?

[night fiber]

Yes

[gilded brook]

So does that mean 1/cos(x)=cos(-x)+tan(-x)sin(-x)?

[night fiber]

Indeed

[gilded brook]

So on paper how would you get there

[night fiber]

You can use the parallel with odd and even numbers

Odd*even is odd, odd*odd is even

[gilded brook]

Just immediately say this is = to this or is there a way to work to that

[night fiber]

With trigonometric functions, parity problems are very common

Here you have -x everywhere so it seems suspicious

And indeed cos is even then cos(-x)=cos(x)

Sin and tan are odd then sin(-x)tan(-x)=sin(x)tan(x)

So the function 1/f is even

And after that, you know tan=sin/cos (essential to know this), and you have sin and cos, so if you remember sin/cos=tan, you want to replace tan by sin/cos

[gilded brook]

So the final answer is sec(x)=f(x)? and then I can do sec^2(x)+sin^2(x)?

To work on the right

Or would it be (sec(x))^2

Or is there no difference to it

[night fiber]

No sec(x)=1/f(x)

Then f(x) =cos(x)

[gilded brook]

Oh

Wait then

That would make the right equation = 0

cos^2(x)+sin^2(x)

And either that's right

Or something went wrong because usually in my experience things never = 0

[night fiber]

cos²(x)+sin²(x)=1

You need to frame this 5 times

With red ink

@gilded brook

To remember this, look at the unit circle I sent above

[gilded brook]

Frame this five times?

[night fiber]

Using pythagoras, you find cos²+sin²=1

I don't speak English

[gilded brook]

Ahh

[night fiber]

But uhh it is maybe the most important formula of trigonometry xD

[gilded brook]

Where is the =1 coming from?

[night fiber]

Using pythagoras, you find cos²+sin²=1

[gilded brook]

Because I'm suppose to evaluate cos^2(x)+sin^2(x) do I still use Pythagoras?

[night fiber]

Cos and sin are the x and y components of a point on the unit circle (circle of radius 1)

Well I guess you can use the result immediately

[old flame]

if you have a right angle triangle of hypotenuse 1, it can be used to find the x and y coordinates of a system with x = 1, at any point on the system, because cos(theta), sin(theta) gives the coordinates

[night fiber]

If you want to prove it, the quick way is to draw the unit circle and to apply the pythagoras theorem

[old flame]

(sorry if I am wrong at any point)

[gilded brook]

So I use that therom to evaluate it?

[night fiber]

If you want to waste paper, derivate cos²+sin², you find something equal to 0 everywhere
That means cos²+sin² is a constant, and cos²(0)+sin²(0)=1

[gilded brook]

Not to solve it but evaluate

[night fiber]

For me, the pythagoras theorem is very appropriate

[old flame]

It can just help you if you get stuck to visualise it

[gilded brook]

Doesn't evaluate mean to break it up into smaller pieces?

[night fiber]

Idk I don't speak English

[old flame]

They mean simplify

Simplify to its simplest form

You dont have to use pythagorus

its just to help you understand

They probably want you to do it algebraically

[gilded brook]

And what is the simpliest form of sin^2(x)+cos^2(x)? Or better question how do you get there

[night fiber]

It is equal to 1 for every x

[old flame]

No I was talking about your original question

Okay yep

[gilded brook]

Yeah

So it is it's simplest form?

[night fiber]

Well you can't be simpler than a constant

[old flame]

so if you look at your original problem

it equals sec(x)

so 1/f(x) = 1/sec(x)

[gilded brook]

1/f(x) =sec(x) yea

No

[old flame]

(1/(sec(x))^2

wait how doesn't it?

[gilded brook]

f(x) =cos(x) I thought

Because cos(-x)+tan(-x)sin(-x)=sec(x)

But those also = 1/f(x)

[old flame]

1/sec(x) is cos(x)

[gilded brook]

So f(x) =cos(x)

[night fiber]

Yeah

[gilded brook]

Wait

Confused

[night fiber]

No you're right

1/f is 1/cos then f is cos

[old flame]

Oh sorry

Am I wrong?

Sorry yeah I think I was wrong

let me double check htat

[night fiber]

Well you were because you forgot the function was 1/f and not f

[old flame]

Yeah sorry I confused you and otld you the wrong thing

Yes you're right @gilded brook

1/sec(x) = cos(x)

oops!

[gilded brook]

Yeah

So Cos^2(x)+sin^2(x) has to be its simplest form

Then why is it asked to evaluate?

[old flame]

so then it is cos^2(x) + sin^2(x)

[gilded brook]

Or simplify

[old flame]

which I think is 1???

[gilded brook]

Yeah it is

[old flame]

so that's the simplification

the answer is 1

and it wants you to show working

that is evaluating

examine/evaluate what this means

[gilded brook]

Alright I'm not good with keywords in math

So I wasn't sure

[old flame]

Oh that's okay

I could help you if you wanted

[gilded brook]

Yeah that be great

[old flame]

Just pm me anytime

[gilded brook]

ight will do

[old flame]

Okily

[gilded brook]

Alright I'm pretty sure I understand the left one but for the right what does for -2<=x<=2 mean in this context

[elder charm]

it's a necessary condition, since arccosine is not defined for values outside of [-1,1]

so they need to specify they don't want you to somehow define it on the entire real axis 🙂

[gilded brook]

So the answer is either -2 inbetween it or 2

Wait but why would it matter if I'm just looking for a simplified version

[elder charm]

So the answer is either -2 inbetween it or 2
Not sure what you mean. If you're confused what -2 <= x <= 2 means, it means (-2 <= x) AND (x <= 2), or, equivalently

$x \in [-2,2]$

[obsidian monolith]

ConfusedReptile:

[elder charm]

that makes x/2 be from -1 to 1, which makes arccos(x/2) defined on that interval.

[terse ravine]

@patent beacon I used the slope formula and found that 1.6 is between 535 and 250 but I don't know what answer it wants.

[gilded brook]

isn't it just 188

[terse ravine]

188 is the f(a)

[elder charm]

h', not h

[gilded brook]

o

[terse ravine]

(402-188) / 0.4 = 535

(88-188) / -0.4 = 250

[elder charm]

392.5 can be a valid answer too. That's (402-88)/(0.8). It depends on how exactly you evaluate the derivative.

the common definition is with f(x+dx), so I guess the 535 one, unless something was said about it

[terse ravine]

535 is wrong I already entered it.

I also entered 535, 250 and was also wrong.

Maybe its asking me to do (535+250) / 2 = 392.5 = 393?

[cerulean pebble]

Why the range of this function f(x)= |x-3|+|x+1| is (-inf,+inf), shouldn`t it be [4, +inf) ?

[elder charm]

Do you perhaps have range confused with domain?

Domain is $(-\infty, \infty)$, range is $[4, \infty)$ indeed.

[obsidian monolith]

ConfusedReptile:

[cerulean pebble]

well...

Mathway btw....

anyway thanks ❤️

Do you know what can be this black points in the graph? @elder charm

[elder charm]

yeah, that's a mistake on their part

Do you know what can be this black points in the graph?
Huh?

[cerulean pebble]

Huh?
@elder charm the first image

it have 5 black points

[elder charm]

no idea, no

[cerulean pebble]

ok, thx

[gilded brook]

Is sqrt (1+16x^2)/(1+16x^2) the final answer for this?

[nocturne dune]

could any1 help me with some precalc hw

[patent beacon]

Ye

[nocturne dune]

I honestly dont even know where to start

like I tried doing it how I normalyl do it but there is no way to check

[uncut mulch]

what are you doing normally

[nocturne dune]

gimme 1 sec ill take a pic

and i have no idea how to do the last 2 tho

[uncut mulch]

you signs wrong for 27

for the rest, apply special ratios

[dreamy summit]

i just like to draw pretty pictures

[uncut mulch]

reference angles / quadrant shifts

[nocturne dune]

wdym by the signs being wrong

[uncut mulch]

is cot positive or negative in Q3?

[nocturne dune]

pos

[uncut mulch]

and what did your answer imply

[nocturne dune]

oh so it would be 5sqrt(39)/39

[uncut mulch]

yes

[nocturne dune]

ahh I see thanks

but btw could u explain the last 2 a bit better because I'm not really too sure how to apply the ref angles and quadrant shifts in that case

[uncut mulch]

also wdym by last 2, there are 4

[nocturne dune]

well yeah I mean the 4

but if u could just help me with 1 for each

i could understand it

[uncut mulch]

the method I would use is to determine the reference angle by applying the inverse function to the abs val of the ratio

[nocturne dune]

ohh I see

ill try that

[uncut mulch]

let that be a,
a = arccos|-1/2|

and do the quadrant stuff using that

and you should be able to get the ones with ratios of -1,0 or 1 without going through all that

[gilded brook]

What is the second solution to tan(theta)+1=0? I got theta=3pi/4 but can't get another one

[uncut mulch]

what's the period of tan

[gilded brook]

It's just it twice?

Period of tan is pi

[uncut mulch]

yes, adding integer multiples of pi and you'll more solutions

[gilded brook]

I'm not understanding what you mean

Ineger multiples so like

+pi

[uncut mulch]

yes

[gilded brook]

So theta=3pi/4+pi is the second answer?

[uncut mulch]

yes

are you only told to give the solutions for 0<=theta<2pi ?

[gilded brook]

Between 0 and 2pi

Can I add pi to 3pi/4 to make it one number

For the answer

[uncut mulch]

yes

[gilded brook]

So 3pi/4 and 7pi/4

Are my two answers between 0 and 2pi

[uncut mulch]

yes

[gilded brook]

ty

[copper breach]

Can someone check my work

I am really confused idk if I solved this correctly

[quaint mason]

I am confused on how to solve for a, b, and c.

[blissful rampart]

I am confused on how to solve for a, b, and c.
@quaint mason use systems of equations

[viscid thistle]

@copper breach 🤔 i would use polar coordinates, but i'm not sure if one teaches about it in pre-university

but:
x = r cos(theta)
y = r sin(theta)

r is given by distance

the distance from (0,0) and (5,5) is 5sqrt(2)

and tg(phi) = 1 # (5-0 = 5 and 5/5 = 1)

then phi = 45° = pi/4

we rotate it counterclockwise 215°

215° + 45° = 260°

so, we rewrite them with respect to it

180 + 80..

we are at 3 quadrant and almost 4

y = -5sqrt(2).sin(80)
x = -5sqrt(2).cos(80), i guess..
when y² + x² = 50², that is our initial "distance point"

y ~ -6,9636424
x ~ -1,22787803

[cerulean pebble]

Help to find the range of g(x)=[(1/x)-1]/[(2/x)+7]

i didn't catch the "1/2" in the answer

my attempt

idk if you can see it btw

[hexed otter]

uhhhh how do u do parabola arch word problems?

[viscid thistle]

Just post it

Someone will

Nw

Read #❓how-to-get-help though

[stone temple]

Uh

you have 10 universal question channels alpha through kappa if i'm not mistaken

And you have more channels here for specific topics

[wintry yacht]

Can anyone help me

[proud raven]

probably

[proud raven]

,rotate -90

[wintry yacht]

That, my friend send me

[proud raven]

which part did you need help with?

[wintry yacht]

Part 1

I solved part 2,3 but part 1 is trixky

[proud raven]

do you understand what that type of function definition is?

[wintry yacht]

Hint?

[proud raven]

like how to figure out how to evaluate f(x)? have you seen a piecewise function before?

[wintry yacht]

Yes but it's tricky

[proud raven]

okay, so looking at part a, can you take a guess at how you'd evaluate that?

im just curious what part of the process you're tripped up on

maybe being more specific, when x = 2, what is f(x)=?

[wintry yacht]

Ah

Ok thanks, i get it now.

[proud raven]

you sure?

thats not the emote of confidence haha

but im happy for you if so

[wintry yacht]

Yes thanks😂

[proud raven]

sure, lmk if you get hung up again, good luck 👍

[willow bear]

don't post across multiple channels please

[viscid thistle]

are repeated and equal roots the same thing, where the discriminant equals 0?

[viscid thistle]

Yea

ok thx

[obsidian monolith]

kin:

kin:

[viscid thistle]

The common ratio is (4-3x)

yea I got to that

It should be <1

yeah I got the correct answer

it's jsut that

I randomly put brackets in U2

Oh then I just did TL; DR

Like U2 is 8-6x(4-3x)

but to make it cancel out with U1

I added brackets to 8-6x so it became (8-6x)(4-3x)

is that wrong to do?

cause for some reason this website is giving me a different result when I put that in

compared to when i put this in

which should be the same thing?

[willow bear]

it's (8-6x)(4-3x) not 8-(6x(4-3x))

[viscid thistle]

oh okay thanks

[hexed otter]

Uhhh if it is fine, please teach me about this, word problems involving conic sections

[wintry yacht]

Hello

the topic is about rational function based on the theorem on the horizontal asymptote

finding the intercepts,domain, horizontal,vertical asymptotes, tables of values and graph

[viscid thistle]

Right

So what do you need help on

[wintry yacht]

I'm troubling about how finding the intercepts, domain and tables of values

[viscid thistle]

Alright

A table of values, do you not know what it is?

[wintry yacht]

Yes I know, about quadratic and polynomial table of values. But not with asymptotes and limit stuff

[viscid thistle]

You are contradicting yourself, didn't you just told me that you needed help with intercepts, domain and table?

A table of values works the same way on a quadratic than on a rational function

[wintry yacht]

Yeah, but not on asymptotes involving limits

[viscid thistle]

Just used to organise important points

Alright as you are not being clear with what you need help on, we'll just keep moving from topic to topic and when you feel okay to continue on your own you let me know

Ok? @wintry yacht

[wintry yacht]

Ok...Uhm can you help me find the domain instead?

[viscid thistle]

Sure

Okay now we are talking

So do you know what a denominator can't have

@wintry yacht can you have the decency of being somewhat active when you are the one asking help?

[wintry yacht]

Not equal to zero?

[viscid thistle]

Yes

[wintry yacht]

Sorry, i only use mobile data and it's raining here. I can't get a good connection rn

[viscid thistle]

A denominator can't be 0, hence whenever our fraction will have a denominator of 0, it will have a problem on domain

Okay you could have said it

I can wait, but you need to tell me

We can keep going whenever you are good

[wintry yacht]

Can i just dm you?

[viscid thistle]

No i prefer here

[wintry yacht]

Ok then, as you said a rational function is defined for all x, except 0. Right?

[viscid thistle]

Depends on the function, but yeah the denominator can't be 0 otherwise it'll be undefined

Bc as you can see, x=0 works perfectly on our function

,w (0²-6)/(0+2)

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Al𝟛dium#4084

[viscid thistle]

In fact that's our y-intercept

[wintry yacht]

Ok. The intercepts of the numerator are the zeros of it?

[viscid thistle]

Intercepts of the numerator?

Why numerator?

The denominator is also a part of the function

[wintry yacht]

I'm just asserting if my own understanding of the definition is correct. Otherwise

[viscid thistle]

Okay

To find any y-intercept, you have to use your logic, the y-intercept occurs when the function hits the y-axis, and where is the y-axis located? At x=0, so whenever you wanna find the y-intercept of a function, all you have to do is plug x=0 and the result will be the y-coordinate of the y-intercept. To sum up, the point of the y-intercept of a function f is (0,f(0))

Same logic applies to x-intercepts

[wintry yacht]

So if f at x=0 as the function says it equal to 2, it's the y-intercept?

[viscid thistle]

Yeah, if you do f(0) on any function f, you'll get the y coordinate of the y-intercept

[wintry yacht]

Then how about the domain?

[viscid thistle]

So if you get f(0)=2, the point where the function hits the y-axis is (0,2)

And now about the domain

A denominator can't be 0, hence whenever our fraction will have a denominator of 0, it will have a problem on domain

We are gonna do the following

We are gonna ask the function where does the denominator = 0, bc the values of x of that eqn will be the ones that make the denominator 0, hence those points will be the ones not on the domain of the function

So all you have to do is equate the denominator to 0 to find the x values that make the denominator to be 0, hence where it'll be undefined

[wintry yacht]

Final solution?

[viscid thistle]

To what

So all you have to do is equate the denominator to 0 to find the x values that make the denominator to be 0, hence where it'll be undefined

[wintry yacht]

What's the domain?

[viscid thistle]

I can't keep repeating myself, it feels like you aren't reading to what i say

Have you set up the eqn

[wintry yacht]

So i just equate the denominator x to zero, and not “approaching”?

[viscid thistle]

Yeah

What i say is what i mean

[wintry yacht]

Ok. Understandable 👌

[viscid thistle]

So how would the eqn look like

???

[wintry yacht]

I don't know

[viscid thistle]

Do you know what equate to 0 means

[wintry yacht]

Undefined

[viscid thistle]

Equate to 0 means = 0

[wintry yacht]

I know, I'm just really confused

[viscid thistle]

And equate the denominator to 0 is just the denominator=0

[wintry yacht]

And then?

[viscid thistle]

I'm here for your doubts

Solve that eqn

Our denominator is x+2

Hence our eqn is $x+2=0$

[obsidian monolith]

Al𝟛dium:

[wintry yacht]

How am i able to solve it, if it's undefined?

[viscid thistle]

You are taking pieces of what i say and then tranform it to other things

The function it's not undefined everywhere, the function is not defined at the points where the denominator is 0, so we are doing this eqn to find the values of x that make the denominator 0

...

That's not what we do here.

If you want just an answer, this is not the place

[wintry yacht]

It just taking to long

[viscid thistle]

So what? Are you gonna keep going through maths without even understanding what you are doing? To then evaluate and have no clue what to do?

[wintry yacht]

No! Not that

[viscid thistle]

Then stop asking for the answer and read what i've said bc i won't repeat myself for the 4th time

[wintry yacht]

Ok, I'll pay attention to whatever you say

[viscid thistle]

Now?

[wintry yacht]

Uhm

[viscid thistle]

The motivation to help you is really flying away from what you are doing.

read everything i've said bc i won't repeat myself for the 4th time

[wintry yacht]

Now as you said. The function is not defined if the denominator equal to zero. And there we can have a clue for what the values of x is?

By setting the denominator equal to zero

[viscid thistle]

Solving $x+2=0$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Yes

[wintry yacht]

Equating

👌

[viscid thistle]

Now it's just algebra

What do you get for x

[wintry yacht]

X+2=0, -2?

[viscid thistle]

Uh yeah x=-2 is the solution for that eqn, hence finally the value of x that makes the denominator to be 0

You can check it yourself

,w ((-2)²-6)/(-2+2)

[wintry yacht]

Ok

Yes

[viscid thistle]

So

Finally

The domain will be all the real numbers except -2

[wintry yacht]

Uhm

[viscid thistle]

Do you know what real numbers means?

[wintry yacht]

👌

Thanks

[viscid thistle]

Please ask if you have any doubts, i can't get in your head to know if you have any.

[wintry yacht]

Nothing anymore. Thanks!👍

[dire bloom]

hi guys

can i ask for help on a question on a test i've already taken

[viscid thistle]

yes you can

[dire bloom]

okay

[viscid thistle]

if you can show proof that it is already over it'd be even better but oh well

[dire bloom]

i took it yesterday but i still have some questions, the teacher doesn't go over tests

its about inequalities with 2 absolute values

[viscid thistle]

sure

go ahead

[dire bloom]

|x^2 - 2x| < |1 - 3x - x^2|

i started out by moving both absolute values to the left side but then i didn't really know what to do next

[viscid thistle]

I would do this by cases

[dire bloom]

oh

so we have to use the definition on both of them?

[viscid thistle]

sadly

[dire bloom]

dang, okay so

[viscid thistle]

it'd be 4 cases in total

[ivory echo]

,rotate

[dire bloom]

oh it's sideways

[dire bloom]

oh

[ivory echo]

I gotchu fam.

[dire bloom]

that bot's cool

[ivory echo]

Say no more :P

Yeee

[viscid thistle]

please make it clearer that it is a one and not another |

[dire bloom]

lol

[ivory echo]

^ just use LaTeX

Everyone should learn it.

High key

[dire bloom]

where do i learn it

i think my brother knows it

[ivory echo]

Well, I can set you off with some basics.

I'mma use DS = $

So if you see DS, that's a dollar sign.

[dire bloom]

ok

[ivory echo]

This is an DS Inline equation DS so it appears normally

Like $3+3$ so.

[obsidian monolith]

Tomstah:

[dire bloom]

i see

[astral mantle]

if you want to typeset latex and edit online it recommend overleaf

[ivory echo]

DSDS Double DSDS DSDS create new lines DSDS

[astral mantle]

i used it for a bit

[ivory echo]

$$3+2+1$$ $$3+3$$

[obsidian monolith]

Tomstah:

[dire bloom]

oh

[ivory echo]

Most unis give overleaf premium for free, mine does.

[astral mantle]

mine aswell

[ivory echo]

Use text outside dollar signs.

[viscid thistle]

there's a file to learn latex at #resources .

[ivory echo]

Anything math goes inside dollar signs.

And then the last big thing is just control sequences.

Which are things led by a \

$\int \sum \mid \leq \geq \neq$

[obsidian monolith]

Tomstah:

[dire bloom]

oh thats nice

ok so how can i make a piecewise

[ivory echo]

https://oeis.org/wiki/List_of_LaTeX_mathematical_symbols

Piecewise is a bit more annoying to do, but:

[viscid thistle]

there's a file to learn latex at #resources .

[ivory echo]

Also use that link ^^^ great reference sheet.

Can I not show him around so I can help him type his problem easier?

[astral mantle]

i think its because this is precalc chat not latex chat

[dire bloom]

oh

where's the latex chat

[astral mantle]

there isnt one

[dire bloom]

ookay

[ivory echo]

Doesn't exist. I'm just trying to show him real fast so hecan type easier.

[dire bloom]

i'll read this document and see

it's fine, i'll figure it out

[ivory echo]

$$T(n) =
\begin{cases}
0 & \text{if $n=1$} \
1 & \text{if $n=2$} \
T(\floor{n/2}) + T(\ceil{n/2}) + 2 & \text{if $n>2$}
\end{cases}$$

[willow bear]

case-sensitive

[obsidian monolith]

Tomstah: