precalculus

[gleaming schooner]

Thanks

[calm whale]

wat

I just told you

it can't be equal to 0

5/x

by definition x cannot be equal to 0

as 5/0

is undefined

[gleaming schooner]

Alright

Why do we have to determine the signs on different intervals?

I don't understand that fully.

[calm whale]

this is a diff method

yes?

the method goes like that

you have the inequality

(x)*(x) > 0

for this to be true

x*x has to be > 0

or -x * -x has to be > 0

[viscid thistle]

@calm whale You don't know what partial fractions are?....

[calm whale]

I mean it doesnt get much simpler than partial fractions

a bit of a terminology barrier

[viscid thistle]

but you said....
"ugh what is a partial fraction 🤔
is that a partial fraction 😄
cause it is just a normal fraction
putting a fancy name infront of it does not make it complicated"

I dont want it to sound complicated ;~; I like partial fractions, they are simple and fun, I have no benefit making them sound "complicated".....why would you think I would do that? ;~;

kinda hurts my feelings

[gleaming schooner]

Anyone around to help with a quick trig function

[viscid thistle]

I'm literally guessing where my question should be but

I'm literally guessing where my question should be but

For the exponential function f(x) C * a^x applies f(0) = 100 and f(10) = 200

Figure out C and a

[viscid thistle]

please @ me if anyone can help me

[serene vortex]

anyone still need help? @gleaming schooner @viscid thistle h

[viscid thistle]

You gotta study

I do

[serene vortex]

ik bro

lmao but im getting it

still just gotta remember the derivative rules and then ill try out the AP worksheet

@viscid thistle whats the 1?

q?

[viscid thistle]

huh?

[serene vortex]

the question

[viscid thistle]

For the exponential function f(x) C * a^x applies f(0) = 100 and f(10) = 200
Figure out C and a

[serene vortex]

f(x)=Ca^x?

[viscid thistle]

I need to fugure out the value for C and a

[serene vortex]

yep

so let me see

ok

so

basically its tellign youy when x=0

ca^x = 100

[viscid thistle]

ohn

so

[serene vortex]

so we can say that ca^0 =100

100=C(1)

[viscid thistle]

wait

[serene vortex]

c=100

[viscid thistle]

if x=0

where do I put everything else that u wrote

I barely understand the math book lol

[serene vortex]

so when x=0

f(0)=Ca^x

Ca^0 =100

a^0=1

C(1)=100

C=100

then just work out a using C when x=10 and f(10)=200

ill brb if u need more help

ok

[viscid thistle]

I understand nothing

[serene vortex]

oh

[viscid thistle]

why is f(0) Ca^x

I hate my teachers so much because they're trash at explaining

smh

[serene vortex]

hop into a voice chat?

[viscid thistle]

uh sure but I might not speak

[serene vortex]

all good

easier for my to speak than type

[viscid thistle]

yeye

oh

idk

honestly

my teacher tells me to write it up in coordinates

like (0, 100) and (10, 200)

yeah I do

in the back of the book

C 100

a^1/10 I have in mind

a = 2^1/10

I don't get them at all

I hope so

[serene vortex]

f(x)=Ca^x

f(0)=100

[viscid thistle]

oh

[serene vortex]

f(10)=200

[viscid thistle]

yeah

[serene vortex]

Ca^0

[viscid thistle]

yeah

you gotta do it for both

yeah

I know

[serene vortex]

c(1)=f(0)

c=100

[viscid thistle]

yeah

yeah kind of

[serene vortex]

f(10)=200

f(10)= Ca^10

[viscid thistle]

yeah

yeah

[serene vortex]

200= 100a^10

[viscid thistle]

does that mean that a^10 is 100?

oh

so I have to divide both

[serene vortex]

2=a^10

[viscid thistle]

yeah

[serene vortex]

2^1/10=a

[viscid thistle]

so let me try to solve this then

so I know that f(x) = C*a^x

and that I have f(0) = 100

If f(0) then it would be just like f(x)

which means that if I have f(0) then I also have a^0

and a^0 is 1

so that would mean that I have 100 * a^0

oh yeah

yup

uh

I'm lost

so I know that a^0 is 1

but idk how to continue

[serene vortex]

f(x)=Ca^x

f(0)=Ca^0

[viscid thistle]

yeah

[serene vortex]

100=Ca^0

[viscid thistle]

yeah

lemme write this down

so what I wrote was

f(x) = C*a^x
f(0) = 100
a^0 = 1
f(0) = Ca^x
100=Ca^0

wait

not ^2

idk how I would write this down though but I assume since it's Ca^0 then it would be 100*1

[serene vortex]

100=C*1

[viscid thistle]

yeah

[serene vortex]

100=C

[viscid thistle]

should I take a pic of my notebook?

[serene vortex]

100=Ca^0

100=C*1

[viscid thistle]

ah

yeah so now I have C

so I have

f(10) = 200

and I know thaat f(0) is 100

oh

[serene vortex]

f(10)=Ca^10

[viscid thistle]

oh I see

[serene vortex]

200=Ca^10

200=100a^10

2=a^10

2^1/10=a

[viscid thistle]

ah

so I have f(10)=ca^10 because f(x) = ^10 if I make f(10)

So now I know that 200 = Ca^10

oh

so 200-100?

[serene vortex]

100*a^10

[viscid thistle]

oh

so

200 = Ca^10

that meanst that 200 = 100*a^10

and I want away the 100's

so 200/100 and 100*a^10/100

so I have 2 = a^10

o ye

so I want away the ^10?

what do I do then

ah

lemme get my calculator

[serene vortex]

2^1/10

[viscid thistle]

well more precisely they said a = 2^1/10 = 1,072

I'm confused what I should do after 2=a^10

probs cause language barrier

nope, Swedish

you could use

the bot

wait

10sqrt?

[serene vortex]

10rt

[viscid thistle]

ah

[serene vortex]

a^1/10

[viscid thistle]

what is the difference between swrt and rt

I mean I've used these things before

but we call it just square root

yeah

well I'm just a highschooler

so

10rt

of a^10

oh

10rt 2=a^10

so that makes the answer to

uhh

2=a^1/10?

[serene vortex]

2=a^10

2^1/10

a

[viscid thistle]

but how do I know what 10rt of 2 is

oh yeah

so 10rt 2=a^10

yeah

so it would be a= 2^1/10

I basicaally switch places for them and proceed to do the rt for both

well I'm gonna try to do another one later

and see if I still have it in my head

well see if it sticks after an hour or something

yeah

nah

next tuesday

I just have a fucked up sleep schedule

yup

yeah

thanks

[serene vortex]

np my b

[serene vortex]

done with my hw, lmk if you need help

[calm whale]

looking at 6

[serene vortex]

wtf

[calm whale]

what :?

[serene vortex]

catesian?

tf

[calm whale]

oh taht simply refers to the x-2/3

thats cartesian

for X

the thing you see in b) is in vector

notation or whatever ists called

I've kinda done this question already

[indigo briar]

Why is the restriction on the inverse function x<0, shouldnt it be x>0??

[calm whale]

the square root

has to always be positive

so when you havea - infront of it

you get x<0

possibly

[frigid hearth]

How do u simplify tan^2(-θ)+1/tan^2(θ)

pls help ;-;

[blazing raven]

name all three pythagorean identities. 😃

[trim hornet]

hello?

[clever inlet]

hi

[trim hornet]

i has a question

[clever inlet]

ok

[trim hornet]

that involves the ambiguous case

i understand that if you have side ac and cb you find height and do what ever

but what if you have side ab and bc

[willow bear]

context?

[trim hornet]

AB=15cm, BC=12cm, and angleA=50 degrees

[willow bear]

is ABC a right triangle, or what?

or is it arbitrary

and most importantly, what are you asked for?

[trim hornet]

how many triangle ar there and no it doesnt specify what kind of traingle it is

are*

[willow bear]

how many triangles there are that have those features?

[trim hornet]

yes

i am sorry if i am making this difficult

arbitrary

[tepid topaz]

Converting (0,1) cartesian coordinates to polar should be (1,pi/4) yes? Book has it as (1, pi/2)

[blazing raven]

no pi/2 I am pretty sure.

Book is right ... once again 😦

[tepid topaz]

Arctan 1 = .785 radians?

Am confused

[blazing raven]

do you know what arctan means?

It means, in this case, what angle gives a tangent of 1.

tan(theta): angle --> ratio
arctan(ratio): ratio --> angle

[tepid topaz]

It should be arctan(1/0) which is undefined, yes? So I'm guessing you get the answer from knowing cos 0, sin 0 is pi/2 radians?

Sin1*

[blazing raven]

no 😃

unit circle. What angle has tan(angle) = 1

[tepid topaz]

Pi/4

Or 45 degrees

[blazing raven]

yup

notice how pi is about 3 and pi/4 should be about 0.75

Arctan 1 = .785 radians? 😃 look familiar?

[tepid topaz]

Yes

[blazing raven]

well there you go.

[tepid topaz]

How did they get pi/2 though

[blazing raven]

because at (0, 1) = (cos(theta), sin(theta)) as long a the original point is on the unit circle.

[tepid topaz]

So its just something you know, not calculate

[blazing raven]

For the special angles kinda.

[gleaming schooner]

Can anyone help me simplify that

I thought that I could simplify (secx)^2 to (tanx)^2 + 1

And then the (tanx)^2 would become 1

But apparently that's wrong, why?

[willow bear]

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

[gleaming schooner]

That's the professor's answer

[signal mesa]

why is "a" not a local maximum?

[viscid thistle]

remember what local extrema is defined as

it needs an open interval (a, b) to have that "c" value inside of it, and then it becomes a relative/local min/max

so in this scenario

your a

has no "neighborhood" or neighbor (sorry i cant remember the terminology) to the left

[signal mesa]

ohh so it's difference for absolute maxima, absolute is just the highest value?

[viscid thistle]

hm

well

absolute/global extrema apply to a specific closed interval too

f can have absolute extrema on [a,b]

(in other words, you look at the endpoints as well)

[signal mesa]

ohh so absolute/global has [] and local is ()

[viscid thistle]

i think so

someone may correct me on that

but idk

seems correct to what i remember

[signal mesa]

thank -u

[timid plank]

@gleaming schooner the answer by the teacher is definetly incorrect, the answer is cot^2(x), you can verify that by substuting an angle for the unsimplified and simplified one

[gleaming schooner]

Yea, that's what I thought lol.

[willow bear]

also if i met your teacher

i would throw tons of at them

[gleaming schooner]

Lol

For that problem, sec^2(x) becomes tan^2(x) + 1, right?

Giving me

[willow bear]

sure that makes sense

[gleaming schooner]

=tex \frac{tan^2(x)+1}{tan^2(x)} - 1

[willow bear]

\tan

:p

[gleaming schooner]

\tan

[willow bear]

=tex \frac{\tan^2(x) + 1}{\tan^2(x)} - 1

[willow bear]

like that

[gleaming schooner]

Alright

Seems to produce a similar output o,o

Ah

I see

Italicized in mine

[willow bear]

=tex a\sin(x) \text{ vs. } asin(x)

[gleaming schooner]

Understood

So wouldn't the two tan^2(x) cancel out?

Or rather, become 1.

[willow bear]

what do you mean

[gleaming schooner]

tan^2(x)/tan^2(x)

= 1, no?

[willow bear]

yes, but

you don't have tan^2(x)/tan^2(x)

you have (tan^2(x)+1)/tan^2(x)

you \textit{can} split it up into $$\frac{\tan^2(x)}{\tan^2(x)} + \frac{1}{\tan^2(x)}$$ though

[gleaming schooner]

Ahhhhh

[willow bear]

but no, you cannot cancel out the a's in (a+b)/a

[gleaming schooner]

So I'd be left with

=tex \frac{1}{\tan^2(x)}

[willow bear]

except that you wouldn't?

you'd have 1 + 1/tan^2(x) - 1

[willow bear]

yup

that's correct

and can be rewritten as cot^2(x) if needed

[gleaming schooner]

Awesome. Thank you!

I've got another one

=tex \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{1 + \cos(x)}

Which gives me

=tex cotx + \frac{\sin(x)}{1 + \cos(x)}

[gleaming schooner]

I'm lost after that though

[quartz garnet]

🤔

How bout trying to reduce them into a common denominator

Seems like it'd become a nice one 👌🏼

[gleaming schooner]

=tex cotx + \frac{\sin(x)}{1 + \cos(x)}

[gleaming schooner]

You mean that?

[patent beacon]

=tex \frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{\sin(x)(1 + \cos(x))}

[patent beacon]

Common denominator

Usually a good first step, but EHH.

[willow bear]

the numerator becomes cos(x) + cos^2(x) + sin^2(x)

= cos(x) + 1

and that can now cancel out

and you're left with 1/sin(x)

@gleaming schooner @patent beacon

[gleaming schooner]

Is that the identity you use?

[willow bear]

yup i used that

[gleaming schooner]

How do you get 1 + cos^2x = sin^2x 😮

[patent beacon]

Oh heh, that's pretty obvious. Shuold have taken it that step further

[willow bear]

i don't

the numerator becomes cos(x) + cos^2(x) + sin^2(x)

the raw cos stays

the other two terms add to 1

[gleaming schooner]

How'd you get that numerator

I guess I'm lost on that

Is this after:

=tex cotx + \frac{\sin(x)}{1 + \cos(x)}

[willow bear]

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

also \cot(x)

=tex \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{1 + \cos(x)} = \frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{\sin(x)(1 + \cos(x))}

[willow bear]

@gleaming schooner do you understand this?

[gleaming schooner]

I didnt know you could cross multiply like that 😮

Since it isn't an equation

[late haven]

it's cause you're multiplying by 1

=tex \frac{\cos(x)}{\sin(x)} \cdot \frac{1 + \cos(x)}{1+\cos(x)}

[late haven]

(half of a cross multiplication)

[patent beacon]

It's a good trick to learn. Easily the only required trick for like 60% of trig identities.

[gleaming schooner]

=tex \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{1 + \cos(x)}

[gleaming schooner]

The problem is ^ though

[willow bear]

@gleaming schooner lolwat i just added the two fractions

=tex \frac{2}{7} + \frac{3}{13} = \frac{2\cdot 13 + 3 \cdot 7}{7 \cdot 13}

[willow bear]

would you call this "cross multiplication"

[gleaming schooner]

No

o,o

[patent beacon]

As was written above

=tex \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{1 + \cos(x)} = \frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{\sin(x)(1 + \cos(x))}

[gleaming schooner]

Alright

I understand

Thanks

I guess I need to brush up on my basic algebra

[quartz garnet]

Yeah...

[gleaming schooner]

I've always struggled with that

Anyway I can get better?

[willow bear]

you do

[patent beacon]

Wow that was ruthless

Even the experts forget some of the important stuff. Don't worry about it. Instead, make sure you understand the process, you don't want to miss out on that

=tex \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{1 + \cos(x)} =
\frac{\cos(x)(1 + \cos(x))}{\sin(x)(1 + \cos(x))} + \frac{\sin(x)\sin(x)}{\sin(x)(1 + \cos(x))} =
\frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{\sin(x)(1 + \cos(x))}

[willow bear]

that got cut off

try a newline

[patent beacon]

Eeh. It looks too scary too

How do new line?

[quartz garnet]

=tex \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{1 + \cos(x)} \
=\frac{\cos(x)(1 + \cos(x))}{\sin(x)(1 + \cos(x))} + \frac{\sin(x)\sin(x)}{\sin(x)(1 + \cos(x))} \
=\frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{\sin(x)(1 + \cos(x))}

[quartz garnet]

Lik dis, mah fella

[patent beacon]

Cool cool, thanks

[quartz garnet]

You need to add \\

(not \)

[gleaming schooner]

Alright, I think I understand a bit better now

I'm gonna get some rest and continue tomorrow

I still need to learn how to verify identities, and double angles and half angles

Oh and graphing trig functions

Gonna be an interesting day tomorrow

Thanks for the help guys

[quartz garnet]

no problem :')

[signal mesa]

If i have a closed continuous function say defined on interval I on domain D, and it opens concave up, are the edges of the function critical points?

[gleaming schooner]

I'm a bit lost on how to simplify this further

[granite stirrup]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[gleaming schooner]

=tex \frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{\sin(x)(1 + \cos(x))}

[willow bear]

@gleaming schooner i told you like twice earlier today though

=tex \frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{\sin(x)(1 + \cos(x))} = \frac{\cos(x) + \cos^2(x) + \sin^2(x)}{\sin(x)(1 + \cos(x))} \ = \frac{\cos(x) + 1}{\sin(x)(1 + \cos(x))} = \frac{1}{\sin(x)}

[willow bear]

surely what i did there does not need explaining?

[gleaming schooner]

Ahh, I didn't think of distributing the cosx

In the first step, how does the top become -1 and the bottom become +1

cosx/cosx = 1

-1/1 = -1

Ah, -1 is -1/1

Wait, is that why?

[willow bear]

ab + a = a(b+1)???

[gleaming schooner]

Oh

The cosx is being factored out

Oops.

After a couple more of these I'll get the hang of it lol

Another question, I'm sorry if this is a dumb one :<

When cosx/cosx cancels out, cosx/cosx doesn't just disappear, it becomes 1.

1 on the top, 1 on the bottom.

[willow bear]

yes, so?

[gleaming schooner]

So, my question is

When (cscx + 1)/(cscx+1) cancel out

The same thing applies

[willow bear]

yes

[gleaming schooner]

Ok, my question was going to be why the top doesn't become 1 + cscx - 1

But

I've got the answer

[willow bear]

what?

[gleaming schooner]

Parenthesis

[willow bear]

the numerator is 1 · (csc(x) - 1)

[gleaming schooner]

Yea

[willow bear]

those are multipliers you have there, not addends

if you had several things being added together in the numerator or denominator of a fraction you would not be able to cancel out any individual addend

the 1 left over after the cancellation is a multiplier

not an addend

[gleaming schooner]

Alright

I just wanna work this one out with you to make sure I'm doing it right:

=tex \frac{1-\sec^2(x)}{\tan(x)-\tan(x)\sec(x)}

[gleaming schooner]

Becomes

=tex \frac{1-\sec(x)\sec(x)}{\tan(x)-\tan(x)\sec(x)}

[gleaming schooner]

Becomes

=tex \frac{1-\sec(x)}{\tan(x)-\tan(x)}

[willow bear]

uhh

what did you try to do there

[gleaming schooner]

I expanded sec^2x

[willow bear]

and why is your denominator 0 all of a sudden 👀👀👀

[gleaming schooner]

secx/secx can't cancel?

Idk :<

[willow bear]

=tex \frac{a+bc}{x+yc} \stackrel{?!}{=} \frac{a+b}{x+y}

[willow bear]

this is what you just did

=tex 1 = \frac{7}{7} = \frac{1 + 2 \cdot 4}{-33 + 10 \cdot 4} \stackrel{?!}{=} \frac{1 + 2}{-33 + 10} = \frac{3}{-23} = -\frac{3}{23}

[willow bear]

this is what you just did

as in, this is the mistake you made

but in your case, as i said, it's even more egregious, since your new denominator is zero!

[gleaming schooner]

I see

Let me re-evaluate

=tex \frac{1-\sec(x)\sec(x)}{\tan(x)-\tan(x)\sec(x)}

[gleaming schooner]

Is what I have

[willow bear]

okay

once again, i'm not really sure where you're going with what you just did, starting from the original expression

[gleaming schooner]

So I shouldn't have expanded? 😮

[willow bear]

i didn't say that, did i?

anyway

if you're simplifying a fraction, you should always try to rewrite the numerator and denominator as products

=tex \frac{1 - \sec^2(t)}{\tan(t) - \tan(t)\sec(t)} = \frac{(1-\sec(t))(1 + \sec(t))}{\tan(t)(1 - \sec(t))} = \ = \frac{1 + \sec(t)}{\tan(t)} = \frac{(1+\sec(t))\cos(t)}{\tan(t)\cos(t)} = \ = \frac{\cos(t)+1}{\sin(t)}

[willow bear]

$$= \cot(t/2)$$, perhaps optionally as that's a slightly lesser known identity ime

[gleaming schooner]

How do you get 1 + and 1 - in the numerator

[willow bear]

...

=tex a^2 - b^2 = (a-b)(a+b)

[willow bear]

you can't not have heard of this

[gleaming schooner]

Ahh

And how'd you go from

=tex \frac{1 + \sec(t)}{\tan(t)}

[gleaming schooner]

To

=tex \frac{1 + \sec(t)(cos(t)}{\tan(t)(cos(t))}

[willow bear]

that's not what i wrote

=tex 1 + \sec(t)(\cos(t)) \neq (1 + \sec(t))\cos(t)

[willow bear]

also

are you asking what i did, or how i knew to multiply and divide by cos(t) rather than something else?

[gleaming schooner]

Where did cos come from

[willow bear]

well sec(t) = 1/cos(t)

and tan(t) = sin(t)/cos(t)

so multiplying the numerator and denominator by cos(t) will clear those "hidden" cos denominators, for lack of a better term

i like to keep things in terms of sin and cos rather than other trig functions

[gleaming schooner]

How do you do (sinx/cosx)*(1/sinx)

Do you cross multiply?

(cosx1)(sinxsinx)

/(cosx*sinx)

[gleaming schooner]

Nvm

I got it 😮

[willow bear]

ahem

fraction multiplication?!

[gleaming schooner]

https://media.discordapp.net/attachments/345008107216306176/380813606910033921/unknown.png

Having trouble with this one.

I got

=tex \frac{(-\sin(x))(-\cot(x))}{\cos(x)}

For the left-hand side.

[willow bear]

uh huh

[gleaming schooner]

Which is:

=tex \frac{-\sin(x)}{\cos(x)} * \frac{-\cot(x)}{\cos(x)}

[willow bear]

no

wrong

[gleaming schooner]

:<

[willow bear]

by your logic, (1*1)/2 would equal 1/2 * 1/2

which it clearly does not

[gleaming schooner]

Ah

[willow bear]

you desperately need to review you're algebra

also, the -1 multipliers cancel out

so your thing is just sin(x)cot(x)/cos(x)

and cot(x) = cos(x)/sin(x), so cot(x) sin(x) = cos(x)

and cos(x)/cos(x) = 1

[nimble oxide]

how can I possibly solve this w/o knowing b? or would I set b=0?

so far I have 1=m(5)+b

But how can I solve for m if they don't give me a y-int.

(asked here since question 1/2/3 were being used)

[willow bear]

there is a formula for the slope of a line connecting two given points

which is also the definition of slope

=tex m := \frac{y_2 - y_1}{x_2 - x_1}

[nimble oxide]

Oh yeah, you're right

[nimble oxide]

m=y

yeah

that

I totally forgot.

-1, thanks 👍🏽

[viscid thistle]

Calculate f(2a) and f(a+1) if f(x) = 3x-2

[viscid thistle]

if someone can help me, please @

[pearl zodiac]

$$f(2a) = 3\cdot (2\cdot x) - 2$$, $$f(a+1) = 3\cdot (x + 1) - 2 = 3\cdot x + 3\cdot 1 - 1 = 3\cdot x + 1$$

[pearl zodiac]

@viscid thistle

[viscid thistle]

so

can you explain

so is it like

[pearl zodiac]

So look at the initial equation 3x-2

Now take the x

And isolate it

So it's 3(x) - 2

[viscid thistle]

yeah

[pearl zodiac]

Now instead of x put what's in the parantheses of the function

[viscid thistle]

oh

2a

3(2a) - 2

[pearl zodiac]

Exactly

[viscid thistle]

so that means that it is 6a

-2

[pearl zodiac]

Yup

[viscid thistle]

then

the same with f(a+1)

[pearl zodiac]

Yes

[viscid thistle]

3(a+1)

which is 3a + 3

[pearl zodiac]

Yes

[viscid thistle]

I see

well that makes more sense now

would be cool if they actually explained that in the book lmao

uh, if f(x) = kx + m then what is f(1) - f(0)

[clever inlet]

well

what's f(1)

?

it's similar to the last question

[viscid thistle]

kx + m?

[clever inlet]

careful

the input is 1

and you substitute it into x

so it makes sense to be:

k(1) + m

or just

k + m right?

[viscid thistle]

I am confused

[clever inlet]

say you have

f(x) = x^2

what would f(2) be?

[viscid thistle]

x^3

I think

no wait

[clever inlet]

so x is what you input into

and 2 is the input

[viscid thistle]

did I say it right?

[clever inlet]

f(2) = (2)^2

[viscid thistle]

oh

[clever inlet]

or just 4

[viscid thistle]

yeah

but I don't see how it relates to f(x) = kx + m

[clever inlet]

so x is what you are inputting into

and for f(1)

1 is the input

[viscid thistle]

yeah

[clever inlet]

so you sub 1 into x

f(1) = k(1) + m

or just

k + m

[viscid thistle]

oh

so the f(x) is the same as k(x)

?

[clever inlet]

f could be really any letter

it's just like a name of the function kind of

[viscid thistle]

well I am confused again then

[willow bear]

f(x) = k * x + m

f(1) = k * 1 + m

@viscid thistle

[viscid thistle]

so

k * 1 + m

• k * 0 - m

that means that it is k

[clever inlet]

yeah pretty much

[timber maple]

Anyone wanna help

[patent beacon]

There's a very simple solution here. Try writing the right as it's actual sum, without sigma notation

[willow bear]

sounds like an application for de moivre's formula 👀

[viscid thistle]

Calculate on the formula y = kx + m on the equation for one line that goes between the points (1,1-1) and (4,2)

[clever inlet]

1-1?

would that just be (1,0)?

[viscid thistle]

I have no idea how to solve this

[clever inlet]

ok

so

[viscid thistle]

I mean I could get k kinda easy right

[clever inlet]

yeah

what'd you get for k?

[viscid thistle]

like it would just be y/x

[clever inlet]

(y2 - y1) / (x2 - x1)

[viscid thistle]

so

(2-1)/4-1)

[clever inlet]

is the first point: (1,1-1) ?

[viscid thistle]

yes

[clever inlet]

wouldn't that just be (1,0)?

[viscid thistle]

?????????

[clever inlet]

cause 1-1

idk

i haven't seen this particular format before

[viscid thistle]

a) k = Δy/Δx = 2−(−1)
4 −1 = 3/3
=1⇒ y = x +m
−1=1+m⇔m = −2⇒ y = x −2

straight out of the answers sheet so it might be weird

[clever inlet]

ok

so i assume the point isn't: (1,1-1)

and is just

(1,-1)

[viscid thistle]

yeah but I can no way solve this on a calculator lol

[clever inlet]

so you have the slope

and a point

[viscid thistle]

syntax all the way

[clever inlet]

just plug in either points

i'll do (1,-1)

so

-1 = 1 + m

m = -2

[viscid thistle]

wait I'm dumb

I think I remember how my teacher did it

[clever inlet]

you just need to find the slope

plug in either points

and solve for m

[viscid thistle]

the steps to go from x1 and x2 = x

the steps for y1 to y2 = y

which is 3/3

[clever inlet]

yeah

i generally do it in one step, but like

it's the same thing

[viscid thistle]

so like

(4,1) 1+3=4

[timid plank]

this doesn't seem pre-calc to me 🤔

[viscid thistle]

I have no idea

I don't know math in English

(-1, 2) -1+3=2

so 3/3

[timid plank]

yes the k value is 3/3 = 1

[viscid thistle]

how do I figure out m

[clever inlet]

plug in one of the points

y = kx + m

[timid plank]

^

[clever inlet]

you know k, and since you have point

you can plug in x and y

and then you just solve for m

[viscid thistle]

so

3=x+m

basically

[clever inlet]

you know x

[viscid thistle]

yeah

[clever inlet]

actually wait

[viscid thistle]

1

[clever inlet]

i don't think either points have y = 3

[viscid thistle]

huh

[clever inlet]

you need to use a single point

[viscid thistle]

yeah (1,-1) and (4,2)

[clever inlet]

let's use (1, -1) cause i think that's easier

[viscid thistle]

okok

[clever inlet]

in that point

y = -1

right?

[viscid thistle]

yeah

[clever inlet]

and x is 1 which you already know

[viscid thistle]

yes

[clever inlet]

and k is 1

y = kx + m

[timid plank]

then just basic algebra

[clever inlet]

(-1) = 1(1) + m

[viscid thistle]

yeah

wait why is it 1(1)

I don't know a lot about how you write it

[clever inlet]

sorry, i thought it would make the substitution more obvious

but anyway

[viscid thistle]

I mean

(-1)=k(1)+m

[clever inlet]

i used it to show the substitution of the know x value

[viscid thistle]

ah

I still don't know how to calculate m tho

[clever inlet]

k is a know value

since we determined the slope

(-1)=1(1)+m

or just

-1 = 1 + m

solve that

[viscid thistle]

I've told you

I don't know

like do I just use basics

[clever inlet]

yes

[viscid thistle]

-1+2=1

m=1

[clever inlet]

-1 = 1 + m

[viscid thistle]

wait no

[clever inlet]

isolate m by subtracting 1 from both sides

[viscid thistle]

??

[clever inlet]

m = -2

[viscid thistle]

I am so confused

I mean

-1=1+m

to make it -1 it's obviously -2 yes

[clever inlet]

yes

[viscid thistle]

is it just like that?

[clever inlet]

yes

it's the value of m that makes the equation true

[viscid thistle]

so if I do it with (1,5 and (3,-3)

[clever inlet]

find k first

[viscid thistle]

I start off with looking at 1,3

to find x

[clever inlet]

ok

[viscid thistle]

1+2=3

x=2

and 5,-3

5-8=-3

y=8

8/3

k=2.66

[clever inlet]

would it not be -8/2?

[viscid thistle]

I am going too fast

yeah

it would

since I go -8

and x=2

-8/2

=4

[clever inlet]

-4?

would it not be negative?

[viscid thistle]

oh yes it would

[clever inlet]

yep

[viscid thistle]

y=-4+m

I can figure out y

by taking out the easiest y value out of the points right

[clever inlet]

it needs to be a single point

[viscid thistle]

can I just pick one?

or no

[clever inlet]

so choose: (1,5) or (3,-3)

[viscid thistle]

5

1,5

[clever inlet]

it doesn't matter which you choose, you'll still get the right answer by the end

so anyway

y = 5

so sub that in

[viscid thistle]

5=-4+m

-4+9=5

y=-4x+9

[clever inlet]

yep

that line is correct

though idk about the working out

[viscid thistle]

working out?

[clever inlet]

you should probably follow a more systematic approach to solving variables

[viscid thistle]

like how

[clever inlet]

5=-4+m

add 4 to both sides

m = 9

same answer

but it's more obvious what you are doing

[viscid thistle]

oh so like

5+4=9

[clever inlet]

is probably not as good

cause the variable just disappears

[viscid thistle]

idk how you would do it then

adding 5+=-4+4

would just make it 9=0+m

[clever inlet]

cause especially when you get to more complicated equations

working out starts being really important

[viscid thistle]

https://i.imgur.com/ZWfVHft.png

I mean this is literally how it is in our powerpoint

[clever inlet]

i mean

that slideshow probably assumes you can already do it

proper working out

[viscid thistle]

idk how I would do it in any other way

[still yew]

Say I have the function $$\frac{x}{x ^ 2+1}$$, what are the holes?

[viscid thistle]

h-holes?...

asymptotes?

[late haven]

holes are the places the function is undefined

in this case, where $$x = \pm \sqrt{-1} $$

[viscid thistle]

@late haven Okay, thats about right, because, Im familiear with undefined areas, just not the term holes, my bad.

Or some some cases

zero divided by zero

We have the function f(x) = 12 - 8x
a) Calculate f(2)

[clever inlet]

2 is the input

and x is what you need to input into

[viscid thistle]

ohhh

so

12-16

-4

f(2) = -4

[clever inlet]

yes

exactly

[viscid thistle]

then how do I solve the equation f(x) = 2

[clever inlet]

for the above given function?

if so:

[viscid thistle]

yeah

[clever inlet]

f(x) = 12 - 8x

and

f(x) = 2

so f(x) is equal to 2 and 12 - 8x right?

12 - 8x = 2

[viscid thistle]

I'm confused

[clever inlet]

basically the output is gonna be 2

and with the function, you need to find x that gives this output

[viscid thistle]

oh

so like

f(2) gave me the output -4

if I am allowed to use decimals

/fractions

[clever inlet]

pretty sure you'll need to

[viscid thistle]

I would do f(1.25)

yeah

12-10

= 2

[clever inlet]

ep

yep

[viscid thistle]

f(1.25) = 2

well thatt was easy

didnt know I wasnt allowed to do fractions in f()

[clever inlet]

fractions are fine

[viscid thistle]

Is it true that f(2) < f(-2)

gonna try to solve that 1

if I did that then the number would be negative wouldn't it

[clever inlet]

f(2) or f(-2)

?

[viscid thistle]

f(-2)

in 12-8x

or would it

negative times negative equals positive but is it really a negative in the first place unless (-8)

[clever inlet]

let's do them one by one

f(2) = 12 - 8(2)

= -4

like previously shown

[viscid thistle]

yeah

ohh yeah

[clever inlet]

f(-2) = 12 - 8(-2)

is?

[viscid thistle]

12-(-16)

[clever inlet]

yeah

[viscid thistle]

which is 28 isn't it

[clever inlet]

indeed

and

• 4 < 28

is obviously true

[viscid thistle]

yes

I think I understand more of math in overall now tbh

[clever inlet]

nice

[viscid thistle]

Prove the difference f(x) < -2

well we knew that f(x) was -4

so it's just -4 < -2

Give an example on
a) An expression(?) with five terms that can be simplified to 8 - 2x

the answer sheet is wrong apperantly

said my fellow book buddy who had the book before me

but

wouldn't it be like

simple just define what x is

[clever inlet]

that's kind of an open ended question isn't it?

[viscid thistle]

I have no idea tbh

b) An fuction so that f(3) = -5

wouldn't it just be done with f(x)=10-5x

[clever inlet]

sounds like another open ended question

[viscid thistle]

I mean I get to choose whatever I want

[clever inlet]

yeah

[viscid thistle]

as long as it works

[clever inlet]

yep

[viscid thistle]

but was I right

Marcus says: The solution to the equation f(x) = 0 is the point where the function y = f(x) crosses the y-axis

Is he right or wrong?

[clever inlet]

should be x axis

[viscid thistle]

why though

I have no idea which one it is tbh

[clever inlet]

x axis is horizontal

y axis is vertical

[late haven]

^

you only cross the y axis at x=0

[empty cove]

Hello.
Could someone help me to translate a mathematical expression into english.
$$\Delta f(x)=f(x+x\small{0})$$
I just started to study calculus and because my english isn't my native sometimes I strugle to find specific articles and learn english calculus terminology.

[willow bear]

did you mean $$f(x + x_0)$$?

[willow bear]

anyway, i don't know of a name for that

[empty cove]

yes.

i meant exaxctly that. but... i did a mistake in formula

$$\Delta f(x_0)=f(x_0+\Delta x)-f(x_0)$$

[twilit zodiac]

https://gyazo.com/d279ed2c9fa3b3f69fb95d8b6f1969eb

is urgent @willow bear how does 3n+2 become 2(3n+2)?

I wont pin u in the future

[willow bear]

=tex \frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad+bc}{bd}

[twilit zodiac]

thank you very much

[viscid thistle]

Uh

Solve the difference graphically and then make an algebraic calculation as a check
a) 5x - 1 < 5 - x
b) 7-4x > 2x - 5

I know how to figure out x both graphically and algebraic

but how the hell do I figure out what the difference should be

should it be x > 1 or should it be x < 1

I have no idea how you figure that out

if you can help me out then please @ me.

[quartz garnet]

What difference?

[clever inlet]

i assume it's a translation of inequality

@viscid thistle which question?

[viscid thistle]

a) 5x - 1 < 5 - x
b) 7-4x > 2x - 5

[clever inlet]

i don't think either question has x > 1 or x < 1

[viscid thistle]

first one is x > 1

[quartz garnet]

no

oh

yeah

wait

no

x'D