#precalculus

1 messages · Page 138 of 1

obsidian monolithBOT
spring thunder
#

,rotate 180

obsidian monolithBOT
rare zephyr
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Wait

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For no.3

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It means if it looks like this right?

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Or nah?

spring thunder
#

the top half would be a reflection about the y axis of the bottom part

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that's just me trying to be overly positive about your graph

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that would be a reflection of, say y=x^2 along y=x

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basically taking the curve of the inverse relation

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about x axis is even simpler

spring thunder
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@rare zephyr rip

slow wharf
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$a_n = n(ln\frac{1}{2} + \sum_{k=1}^n ln\frac{(k+1)^2}{k(k+2)}$

obsidian monolithBOT
slow wharf
#

I am trying to solve this problem

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I know about the trick that that is equivalent to

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$a_n = n(ln\frac{1}{2} + ln\prod_{k=1}^n \frac{(k+1)^2}{k(k+2)})$

obsidian monolithBOT
slow wharf
#

Now the product

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The first few elements of it are

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$\frac{2^2}{1 \cdot3 }\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot 5}\cdot\frac{5^2}{4\cdot 6}$

obsidian monolithBOT
slow wharf
#

We can cross them out

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$\frac{2^{\cancel{2}}}{1 \cdot\cancel{3}}\cdot\frac{\cancel{3}^\cancel{2}}{\cancel{2}\cdot\cancel{4}}\cdot\frac{\cancel{4}^\cancel{2}}{\cancel{3}\cdot \cancel{5}}\cdot\frac{5^\cancel{2}}{\cancel{4}\cdot 6}$

obsidian monolithBOT
slow wharf
#

Well you understand the pattern

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So the result of the product would be $2*\lim\frac{k+1}{k+2}$

obsidian monolithBOT
slow wharf
#

Which is just 2

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So a_n is

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$a_n = n(ln\frac{1}{2} + ln2)$

obsidian monolithBOT
slow wharf
#

And this is supposed to be $n\cdot 0$

obsidian monolithBOT
slow wharf
#

But I am somewhere wrong...

spring thunder
#

that thing is telescopic af indeed

slow wharf
#

Yes

spring thunder
#

lemme just make sure i don't say shit

slow wharf
#

I was able to solve it

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And wow it is amazing

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I guess when you get $\infty\cdot0$ there is always another way to solve it?

obsidian monolithBOT
spring thunder
#

yes

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the 2 * lim (k+1)/(k+2) you did was p much an illegal step

slow wharf
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illegal step?

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Why?

spring thunder
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and i was searching for errors in your pattern lol

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well you had 0*inf as you said

slow wharf
#

There are no errors, there's just another way of doing it

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You want me to post it?

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Well

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Do you know that

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$ln(a) + ln(b) = ln(ab)$

obsidian monolithBOT
spring thunder
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oml i don't megathink

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jk

slow wharf
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Well

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xD

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I did this

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I turned this

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Um...

obsidian monolithBOT
slow wharf
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$n(ln\frac{1}{2}+ln\prod_{k=1}^n\frac{(k+1)^2}{k(k+2)})$

obsidian monolithBOT
slow wharf
#

This

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into

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$nln(\frac{1}{2}\cdot \prod_{k=1}^n\frac{(k+1)^2}{k(k+2)})$

obsidian monolithBOT
slow wharf
#

And solve it from there

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The inside of the ln turns out to be a nice product that you can cancel out very well

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You get $\frac{k+1}{k+2}$

obsidian monolithBOT
slow wharf
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$a_n = nln\frac{n+1}{n+2}$

obsidian monolithBOT
slow wharf
#

Now does that look familiar?

spring thunder
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not familiar enugh ig

slow wharf
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You sure?

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$ln(\frac{n+1}{n+2})^n$

obsidian monolithBOT
slow wharf
#

Looks more familiar now?

spring thunder
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no i see dat e indeed

slow wharf
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😃

spring thunder
thick raptor
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$$\ln\left[\frac{(k+1)^2}{k(k+2)}\right]=[\ln(k+1)-\ln(k)]-[\ln(k+2)-\ln(k+1)]$$

obsidian monolithBOT
thick raptor
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that sum telescopes pretty directly FYI @slow wharf

slow wharf
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wut

thick raptor
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ur sum from earlier

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the sum telescopes

remote musk
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Is this precalculus material? I wonder why I'm not good at it

open apex
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Technically, although it's more calc then pre-calc

viscid thistle
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yes, this is defintely claculus

jagged epoch
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I actually have that stuff in my pre-calc class

slow wharf
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I am able to find the limit of a_n

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It is $\frac{1-c}{\sqrt{e}}$

obsidian monolithBOT
slow wharf
#

But I am having trouble finding the limit of b_n

royal gull
#

Maybe stolz theorem or sth

serene heath
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yikes that looks nasty

slow wharf
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Yes ._.

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Wolframalpha doesn't seem to know how to solve it

royal gull
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Try stolz cesaero it might work

slow wharf
#

I have never heard of that

royal gull
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Its a useful theorem that helps determine limit of a fraction

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Easy in use as well

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Yeah thats gotta be it

slow wharf
#

Okay, I try

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It is not it..

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I tried usign it

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On just the top one

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and it turns out to be 0

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@royal gull

serene heath
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the numerator of the first one is 1/e

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no 1/e^2 actually

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but the denominator

slow wharf
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What??

serene heath
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lol na its 1

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ignore me

royal gull
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@slow wharf top one? You need to use it on entire fraction, because there are two sequences: one in nunerator one in denominator (you have to like separate it)

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B_n= c_n/d_n and you need to check limit of c_(n+1)-c_n / d_(n+1)-d_n

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Im on Phone sry but you get it right?

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I thought you want to work out limit of bn first then do an/bn, right?

slow wharf
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Wait, let's do it on a simpler task

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This is the same one but with different numbers

royal gull
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No its easy i think you Just used stolz cesaero wrongly

slow wharf
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Can I first find the limit of a_n+1

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The limit of a_n

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Then limit b_n+1

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I should be able to

royal gull
#

?

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First line

slow wharf
#

What's that

royal gull
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Limit of this is limit of your sequence B_n using cesaero theorem

slow wharf
#

Is it not this?

royal gull
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Ye but youre supposed to do it on YOUR sequence bn

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Let me write it since there are too many same names for sequences

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Your oroginal B_n can be divided on two sequences like that

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And at the very bottom is the stolz thereom in use

slow wharf
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I am giving up...

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It is tooo long

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There must be a simpler way

royal gull
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hmm I thought it would cancel some things out but if not then yeah, there might be some calculations, if it doesnt give a nice ersult then there should be an easier way

slow wharf
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<@&286206848099549185> Any idea? 😭

royal gull
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repost pic

slow wharf
royal gull
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ohh this one

slow wharf
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This one seems a little bit simpler

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But the same thing applies

royal gull
#

a_n looks very much like three sequence theorem (if thats how its called in english)

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basically bound a_n frrom left and right, from left for example by cn/sqrt(n^2 +n)

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and from right by cn/sqrt(n^2+1)

slow wharf
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hmm

royal gull
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these boundary sequences have same limit? if yes then by the squeeze theorem an has the same one since its inbetween

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yeah, the limit of a_n is c

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Squeeze theroem!

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thats what its called

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@slow wharf do you know why I used these sequences to buond our a_n? Or how I found them?

slow wharf
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No ._.

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That is supposed to be

royal gull
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so basically the smaller boundary is just c/sqrt(n^2 + n) added together n times

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its the smallest part of your sequence

slow wharf
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$a_b = \sum_{k=1}^{\infty} \frac{c}{\sqrt{n^2+n}}$

obsidian monolithBOT
slow wharf
#

Right?

royal gull
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yeah

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for the uppoer bound you take the smallest denominatior

slow wharf
#

a_n, but ok

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$\frac{c}{\sqrt{2}}$

obsidian monolithBOT
royal gull
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wait no

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your summation is wrong

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on the pic

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you dont add up till infinity

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yo go from n^2 +1 in denominator to n^2 +n, meaning n terms

serene heath
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random k btw

slow wharf
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it goes to infinity

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No

royal gull
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no, you stop on denominator n^2 + n

slow wharf
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Rephrase that

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It converges, no?

royal gull
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yes it does

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to c

slow wharf
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I understand that if it were

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$\frac{1}{\sqrt{n^2+n}}$

obsidian monolithBOT
slow wharf
#

It would converge to 1

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So it converges to what is on the top?

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Well okay, so a_n = c

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b_n

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should be

royal gull
slow wharf
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$b_n = \frac{\frac{n(n+1)}{2}}{c(n+2)}-\frac{n}{2c}$

obsidian monolithBOT
slow wharf
#

That ends up being $\frac{-n}{2c(n+2)}$

obsidian monolithBOT
slow wharf
#

so $\lim \frac{c}{\frac{-n}{2c(n+2)}}$ i end up with $2c^2-8=0 but it is incorrect...$

obsidian monolithBOT
royal gull
#

@slow wharf check out my pic, in such examples its easy to use that, if your sequence is between two sequences that converge to the same value then you roriginial sequence also converges to the same thing (cause its in between, squeeze theoreM)

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Ok i need to leawrn latex, but I think your calculations are wrongh?

slow wharf
#

wrong?

royal gull
#

what is the right answer?

slow wharf
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1/4

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Oh

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it b_n/a_n

royal gull
#

yeah I got c=1/4

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yeah you did upside down lol

dim citrus
#

@tawdry elbow iq

tawdry elbow
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🆑🅰🅿🅿ℹ🆖

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Omg it’s 3blue1brown

idle dust
#

REEEE

unique nacelle
#

It is estimated that the cost of computers and computer equipment is decreasing at an effective rate of 9% per year. If this rate is maintained, how long till it take computer costs to decline by one third

open apex
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Some help would be mucho appreciated

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👍🏻

tender roost
#

ping me back for answers

tender roost
#

nvm i just got answer from a friend

dim charm
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@open apex

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this seems famat-y

open apex
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Hmst

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It is

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Why are you wondering

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@dim charm

dim charm
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idk

open apex
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Seems odd, only people from Florida call it FAMAT

dim charm
#

yep

errant yew
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Hey guys

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I'm having trouble with this one math problem could anyone please help

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if a b and c are digits and ab x cb = ddd determine the sum of ab+cb
a) 49 b) 57 c) 64 d) 72 e) 80

unique nacelle
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solve 5=3^x

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log_3(5)=x right?

serene heath
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yes

unique nacelle
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ok but how do you solve on calculator?

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it only does base 10 ?

serene heath
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id leave it exact

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or you can use the base change formula to write it in base 10

unique nacelle
#

ok so then for 7=5^(x+4)

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log_5(7)-4=x

serene heath
#

yes

unique nacelle
#

ok ty

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wait how about when exponent is negative

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nvm its just neg log

slow wharf
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$\sqrt{x\sqrt{x}}=(x\cdot x^{\frac{1}{2}})^{\frac{1}{2}} = (x^{\frac{3}{2}})^{\frac{1}{2}}$

obsidian monolithBOT
slow wharf
#

But that is not

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$x^{\frac{3}{4}}$

obsidian monolithBOT
slow wharf
#

Or is it?

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Wait, of course it is, why is this confusing me

serene heath
slow wharf
#

$x^{\frac{1}{2}}^{\frac{1}{2}} = (x^{\frac{1}{2}})^{\frac{1}{2}}$

obsidian monolithBOT
remote musk
#

it's just x^0.5 * x^0.25 = x ^(0.5 + 0.25)

idle dust
#

hi

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can someone just quick help me with this

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i forgot how to do it

slow wharf
#

@idle dust Convert the number -512 into polar form

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If you get the cube root of both sides you will get that z is the cube root of the polar form of -512

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The cube root of a complex number will yield 3 possible solutions.

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And you have to find the one that's between 270 and 360 degrees

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You following?

spring thunder
#

✝

idle dust
#

J

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ok

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so

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-512 is 512(cos 180 +0)

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yes

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i am following

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oK

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so the solutions change by 120 degrees each right

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so the next solution would be at 300 degrees

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soooo i need to find rectangular form

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oh i know

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cos is x and sin is y

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ok ok i got this

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8cos 300 +8isin 300

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right

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@slow wharf is that right?

spring thunder
#

he certainly sleeping ig

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you forgot to 'cube root' the argument tho

idle dust
#

o

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ok

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its 8

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right

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the distance is 8

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or is it a vector???

spring thunder
#

i'm ok for the 8 np with that

idle dust
#

its technically a vector right?

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ok thank

spring thunder
#

you forgot about the angle tho

idle dust
#

the angle 300

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degree

spring thunder
#

it ain't

idle dust
#

huh

spring thunder
#

512 e^(i*pi)

idle dust
#

ok? but i dont need to use exponential form

spring thunder
#

one cube root of it is 8 e^(i* pi/3)

idle dust
#

oh

spring thunder
#

you don't have to

idle dust
#

uhh

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so

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do i divide the angle by 8

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so it becomes 180/8 +120/8 k

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k is random integer?

spring thunder
#

why by 8?

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it's a 3 i divided by

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(180+360k)/3 ° : those are your possible angle sols

slow wharf
#

I was sleeping lol

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You have a formula fir cube root

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Gimmie a sec

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$512^{1/3}(cos(\frac{\frac{\pi}{2}+2k\pi}{2})+isin(\frac{\frac{\pi}{2}+2k\pi}{2})), k=0,1,2$

obsidian monolithBOT
slow wharf
#

Those should be the three answers

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@idle dust

idle dust
#

oH

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omg

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sorry i was taking a dump

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lol

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you guys are geniuses

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right 60°+120k

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so 300 still

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bruh

slow wharf
#

Howd you get that

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Oh sec

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I wrote wrogg

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Wrong

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$512^{1/3}(cos(\frac{\frac{\pi}{2}+2k\pi}{3})+isin(\frac{\frac{\pi}{2}+2k\pi}{3})), k=0,1,2$

obsidian monolithBOT
slow wharf
#

Wait

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Your angle is not pi/2 it's 0 lol

rocky bison
slow wharf
#

$512^{1/3}(cos(\frac{2k\pi}{3})+isin(\frac{2k\pi}{3})), k=0,1,2$

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Damnit again

obsidian monolithBOT
slow wharf
#

There you go

rocky bison
#

\left( and \right)

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Makes the brackets scale

slow wharf
#

Ohhhhh

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That's what it meant, damn.

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Thanks @rocky bison

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Something's wrong...

rocky bison
#

What's happening?

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What are you trying to do?

slow wharf
#

Solving frank's problem

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The angle is /pi/2

rocky bison
#

pi/2 is not between 270 and 360

slow wharf
#

$512^{1/3}\left(cos\left(\frac{\frac{\pi}{2}+2k\pi}{3}\right)+isin\left(\frac{\frac{\pi}{2}+2k\pi}{3}\right)\right), k=0,1,2$

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You said it was \left and \right

rocky bison
#

Yea

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\left(

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Not just \left

slow wharf
#

Oh

rocky bison
#

You can kinda use your intuition of complex numbers to eyeball a solution

obsidian monolithBOT
rocky bison
#

The argument of the product of two complex numbers is the sum of the respective number's arguments

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So arg(z_1*z_2)=arg(z_1)+arg(z_2)

slow wharf
#

We are not talking about product, we are talking about root, no?

rocky bison
#

What is a cube

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is z * z * z

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So the product

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of three z's

slow wharf
#

Ahh

rocky bison
#

And -1 (We can ignore the magnitude given we're just looking at the angle) is an angle of pi

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,$ \arg(z^3)=\pi+2n\pi

obsidian monolithBOT
rocky bison
#

We can ignore the 2npi for now tho tbh

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Just to keep things simple

slow wharf
#

n is 0,1,2

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Right

rocky bison
#

n is any possibl natural number

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Integer even

slow wharf
#

hmm

rocky bison
#

-pi is the same rotation

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3pi is the same rotation

slow wharf
#

yeah

rocky bison
#

So any integer works

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So what possible solutions do we have?

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pi/3 is one

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pi/3 + pi/3 + pi/3 = pi

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But that's not in the range we're given

slow wharf
#

Sec

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-pi * -pi * -pi

rocky bison
#

Urm no

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The angles add up

slow wharf
#

3pi * 3pi * 3pi?

rocky bison
#

You're multiplying the arguments

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Which isn't what happens when you multiply two complex numbers

slow wharf
#

Oh, you add them?

rocky bison
#

yes

slow wharf
#

So 3pi + 3pi + 3pi and 5pi + 5pi + 5pi

rocky bison
#

First lets check that they're correct numbers

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So first one gives 9pi

slow wharf
#

They are

rocky bison
#

Does that have a value of n

slow wharf
#

That's still pi

rocky bison
#

Yes

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Now is it in the range given?

slow wharf
#

9pi is not, that's still just pi

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You said if it has a value of n

rocky bison
#

yes

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so it is correct

slow wharf
#

It's pi + 2(4)pi

rocky bison
#

but they're giving you a range

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there's infinitely many solutions

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It's asking for a specific solution(s)

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So you're saying the argument of z is 3pi

idle dust
#

hey wahtcha guys talkin abuot

rocky bison
#

Let's convert our range and say our argument is theta

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So we're told

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,$ 270<\theta<360

slow wharf
#

@idle dust The quiz you posted

obsidian monolithBOT
rocky bison
#

Convert this to radians for ease

idle dust
#

oh

#

ok i saw

rocky bison
#

,$ \frac32\pi<\theta<2\pi

obsidian monolithBOT
idle dust
#

yesh

rocky bison
#

So, is your chosen value of theta within this range?

idle dust
#

yes

#

is 300

rocky bison
#

He said 3pi

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Which is not 300

idle dust
#

3pi is 270

rocky bison
#

What

idle dust
#

bruh

#

but the range is 60 ° +120k°

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if k is 2

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then it becomes 300

#

OH WAIT

slow wharf
#

3pi is 540

#

No?

idle dust
#

IT STARTS AT

#

180

#

oK

rocky bison
#

We're also working in radians

idle dust
#

w8t a minute

#

no it doesnt

#

uhhh

rocky bison
#

If we start darting inbetween radians and degrees I'm going to confuse myself

idle dust
#

it was 180 +360k

rocky bison
#

So let's pick one

idle dust
#

so i divided by 3

#

ohh

rocky bison
#

Do we want radians or degrees

idle dust
#

ok no im not wrong

#

it is 300

#

i want degrees

slow wharf
#

$512^{1/3}\left(cos\left(\frac{\pi+2k\pi}{3}\right)+isin\left(\frac{\pi+2k\pi}{3}\right)\right), k=0,1,2$

rocky bison
#

I never said you were wrong.

obsidian monolithBOT
idle dust
#

3pi radians is 540

rocky bison
#

Please stop posting

#

That's a horrendous overcomplication imo

slow wharf
#

It should be correct now

idle dust
#

ohk?

slow wharf
#

Well, I used less of my brain when solving it like that so...

idle dust
#

bruh

rocky bison
#

I think solving things intuitively is better tbh

idle dust
#

yeah me too

#

now im just confused

#

anywae

slow wharf
#

-_-

idle dust
#

lol its fine i think i found the answer

slow wharf
#

It's 5pi/3, right?

idle dust
#

yo

#

brother

#

i plugged sin of 300 degrees and it is the same thing as ur stuff

#

so iw as correct

#

the argument is 300

void coral
#

I have a quick question

#

Thats the factored form of that cubic function

#

The a value is in front of the x that is outside the brackets right?

idle dust
void coral
#

Looking at the factored

slow wharf
#

What do you mean @void coral

#

That's the same as

#

$(x)(2x-5)(x+4) = (2x-5)x(x+4) = (2x-5)(x+4)x$

obsidian monolithBOT
void coral
#

Oh I just answered my own question

idle dust
#

what a value

void coral
#

Thanks lol

idle dust
#

bruh ok bye

#

i have no clue wahts going on

#

;-;

slow wharf
#

What's going on is that it's already 2am and I am in my bed talking about math

#

I will try to force myself to sleep now, good night

void coral
#

Looks like it was time to sleep a while ago

slow wharf
#

Yes ._.

unique nacelle
alpine basin
#

Hi, would the answer to the inequality: (3x + 1) (5-10x) > 0 be (-1/3, 1/2) in interval notation?

fading token
#

Yep

#

You also have to mention the linear terms can't both be negative simultaneously

#

But other than that, you're good

slow wharf
#

Yes and no

#

Those indeed are null points

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but

#

You have to check the sign

#

As far as I see it is going to be

#

$D_f = \left(-\infty, -\frac{1}{3}\right)\bigcup\left(\frac{1}{2},+\infty\right)$

obsidian monolithBOT
slow wharf
#

Actually no

#

Opposite

#

From -1/3 to 1/2, open interval

fading token
#

:p

tulip siren
#

was wondering if there were any good resources for precal in general. just switched majors from english, doing math for the first time in over five years and I completely failed my first precalc exam full of things I used to know, I'm unbelievably lost now and I need a way to recover

fading token
#

I would personally recommend Schaum's Outlines first, and some additional more topic-specific resources if needed

tulip siren
#

thank you, i'll take a look at that book

graceful egret
#

find f'(x)

#

f(x)=5/7x^4

viscid thistle
graceful egret
#

the answer is -20/7x^5

#

can someone explain how?

viscid thistle
#

yeah

#

so the coefficient is (5/7) * (1/x^4)

#

whats derivative of 1/x^4

#

bring it up to, that's x^-4

#

n*x^(n-1)

#

bring down the -4

#

thats -4(x)^(-5)

#

-4/x^5

#

then bring back the 5/7

graceful egret
#

ok my slow brain needs to break this down one sec...

#

ok so

#

are you saying i bring the 7x^4 to the numerator?

#

then get the derivative of that

viscid thistle
#

no

#

,w graph 5/(7x^4)

obsidian monolithBOT
viscid thistle
#

completely ignore the coefficient of 5/7

#

then all that's left is 1/x^4

#

right?

graceful egret
#

ok sure

viscid thistle
#

so whats the derivative of 1/x^4

graceful egret
#

1/4x^3?

viscid thistle
#

first you have to bring the variable x to the denominator

#

that becomes

#

x^-4

#

agree?

graceful egret
#

its already in the denominator

viscid thistle
#

numerator

#

bring it to numerator

graceful egret
#

okok

viscid thistle
#

that is x^-4 right?

graceful egret
#

yes

viscid thistle
#

So the exponent rule says

#

the derivative of x^n = n * x ^(n-1)

#

So bring the current power down as a muliplier

#

and take away 1 unit

graceful egret
#

ok gotcha

viscid thistle
#

x^-4

#

where n = -4

#

what do you get

graceful egret
#

-5

#

^-5

viscid thistle
#

whats the whole answer then

#

whats derivative of x^-4

graceful egret
#

-4x^-5

viscid thistle
#

SIMPLIFY that

#

-4/ what

graceful egret
#

well i know the answer

#

but i dont get how to get to it from this point

#

so now we have

viscid thistle
#

bring the variable back down

#

-4x^-5

#

simplify that

#

show me

graceful egret
#

5*-4x^-5/7

#

thats what we are at right?

#

20x^-5/7

viscid thistle
#

right, but just to keep stuff simple ignore the 5/7 for now

#

we'll get to that

graceful egret
#

oh ok

viscid thistle
#

so simplify -4x^-5

#

what do you get

#

bring the variable back down

#

since its a negative power

graceful egret
#

only the variable?

viscid thistle
#

and the power

graceful egret
#

-4/x^-5

viscid thistle
#

but positive now

graceful egret
#

oh yea

viscid thistle
#

1/x = x^-1

#

= 1/x^1

#

ok fix it show me thonker

graceful egret
#

4/x^-5

#

?

viscid thistle
#

positive exponent

#

1/x^1 = x^-1 = 1/x^1

#

you're messing up the powers here

#

if you got 1/x^2

#

thats like x^-2

graceful egret
#

oh ok

#

thats helpful

viscid thistle
#

you got x^-5

#

thats like 1/x^5

#

simplify

#

-4x^-5

graceful egret
#

1/4x^5

viscid thistle
#

only the variable goes down

#

and the 4 is -4

#

-4 / what

graceful egret
#

x^5

viscid thistle
#

write it

graceful egret
#

-4/x^5

viscid thistle
#

yee

#

so -4/x^5

#

what coefficient did we ignore for now?

#

do you remember

graceful egret
#

5/7

viscid thistle
#

so what is 1/2 * 1/2

graceful egret
#

1/4

viscid thistle
#

so you multiply across the fraction right?

#

1 *1 / 2 * 2

#

right?

graceful egret
#

thats just 1?

viscid thistle
#

right but showing the steps

#

1/ 2 * 2

#

thats 1/4

#

just muliplied across, you see that?

graceful egret
#

ok yea

viscid thistle
#

So if we got

#

5/7 * -4/x^5

#

just muliply across

#

multiply

#

thats (5)(-4)/(7 * x^5)

#

simplify that

graceful egret
#

-20/7x^5

viscid thistle
graceful egret
#

haha

#

thanks

#

quick question

viscid thistle
#

yea

graceful egret
#

do you enjoy doing math>

#

or helping people

#

or both

viscid thistle
#

Probably a mix of both

graceful egret
#

so what we just did there

viscid thistle
#

It's rewarding to see people learn

graceful egret
#

did you actually enjoy that?

#

be honest haha

viscid thistle
#

To a degree it's frustrating, but I feel like you probably understood enough of it to finish it

#

practice this one

graceful egret
#

hahaha yea i will

viscid thistle
#

7/(5x^3)

#

whats deriv of this

graceful egret
#

i was flying through the other derviatives but that one had me so stuck

#

give me a minute....

viscid thistle
#

whats the coefficient we ignore for now?

#

start with that

graceful egret
#

7/5

viscid thistle
#

yeah, so whats left?

graceful egret
#

1/x^3

#

which is like

#

x^-3

viscid thistle
#

keep goin fren

graceful egret
#

OMG IM LEARNING

viscid thistle
#

now do the nx^(n-1) bit

graceful egret
#

-3x^-4

viscid thistle
#

yup now bring in coefficient

graceful egret
#

what do i do for next step?

viscid thistle
#

well, bring the power back down anyways

#

-3x^-4

#

fix the power

#

what do you get

graceful egret
#

-3/x^4

viscid thistle
#

now bring in what you ignored at the start

graceful egret
#

7/5

viscid thistle
#

same stuff, mulitiply across

graceful egret
#

-21/5x^4

viscid thistle
graceful egret
#

awesome thanks so much

viscid thistle
#

,w derivative (7/(5x^3))

obsidian monolithBOT
graceful egret
#

how long will you be here?

viscid thistle
#

ping me if needed, or dm if im here ill help. If not wait 15 minutes and do @ helpers

graceful egret
#

ok, ill make sure i try all my ideas then ill ask

#

probably will need some help later tonight

graceful egret
#

im just plugging 13 into x and getting 0/0 as my final answer.

#

this is clearly not how i should be solving this problem

#

any help?

viscid thistle
#

can anyone help me with a few problems

#

it would be much apprectiated

graceful egret
#

omg you didnt read the rules!!!angerywoog

viscid thistle
#

😯

graceful egret
#

-Don't ask "Can I ask a question?" The answer is always yes.

#

im going to have to call the police on you

viscid thistle
#

SHIT

#

BRO IM SORRY

#

plz no

#

can u help me with derivitives of ln and e

graceful egret
#

i definitely can not

#

currently waiting for help

#

and waiting 6 more minutes to ping helpers

finite iris
#

@graceful egret factor each quadratic into (x - a)(x - b)

graceful egret
#

i dont understand

finite iris
#

x^2 - 2x + 1 = (x - 1)(x - 1) right?

#

do the same with x^2 - 169
and x^2 + 3x - 208

graceful egret
#

yes

#

ok

#

1 sec

#

so i need 2 numbers that add up to -208 and multiply to 3 right>

#

?

#

wait

#

other way around i mean

#

so i get -16 and 13

#

i kinda forgot what to do next

#

ok so i got x^2+16x-13x-208/(x+13)(x-13)

#

what now @finite iris

finite iris
#

factor the numerator more

graceful egret
#

how

finite iris
#

@graceful egret write x^2 + 3x - 208 as (x + a)(x + b)

#

if you don't recall how to do this, google how the factor quadratics

graceful egret
#

you said factor the denominator more?

finite iris
#

sorry, I meant numerator

graceful egret
patent beacon
#

That is a line that passes through (3,0) and has the same slope as x³ - 9x at x = 3

#

A natural question is... What is that slope?

#

@graceful egret

graceful egret
#

not sure

#

obviously something to do with the x^3-9x

patent beacon
#

Know how to take a derivative?

graceful egret
#

yes

#

3x^2-9

patent beacon
#

Exactly. That tells you the slope at x

#

We want the slope at x = 3

graceful egret
#

so plug in 3?

#

to x

patent beacon
#

3(3)² - 9 = 18

#

So that's the slope of x³ - 9x at x = 3

graceful egret
#

so the equation is...?

patent beacon
#

What's the equation of a line that has slope 18, and passes through (3, 0)?

#

y = 18(x - 3)

graceful egret
#

18x?

#

ohhhh

patent beacon
#

y = m(x - a) + b
is a line with slope m, that passes through (a, b)

#

Useful form

graceful egret
#

GOTCHA

#

thanks

#

i forgot the x-a

#

got time for another question?

patent beacon
#

,ask graph x³ - 9x, 18(x - 3)

obsidian monolithBOT
patent beacon
#

Very nice

#

Sure ask away

graceful egret
#

ok hold on, need to see what i need most help on for my test tomorrow haha

#

ok truthfully i havent even tried this one yet but id rather get the help before you leave

patent beacon
#

"Tangent line is horizontal" means the slope of the tangent line is zero, means the derivative is zero

#

The question is asking "When is the derivative of 5x² - 2x + 4 equal to 0?"

graceful egret
#

so i feel like i could get that by plugging in numbers and testing

#

but there has to be a better wat

#

way

patent beacon
#

Waste of your time it's easier than that

#

Natural question, what's the derivative?

graceful egret
#

10x-2

#

oh

#

x=.2?

patent beacon
#

Yussir

graceful egret
#

HAHA LEARNING 😄

patent beacon
#

10x - 2 = 0
x = 1/5

#

,ask graph 5x² - 2x + 4

obsidian monolithBOT
patent beacon
#

You can see the flat point at x = 0.2 there

graceful egret
#

YES

#

so

#

The​ point(s) at which the tangent line is horizontal is​ (are)

#

(.2,4)

#

?

patent beacon
#

Oh yes you need the y-coord

graceful egret
#

i put that in and it said it was wrong

patent beacon
#

y = 5x² - 2x + 4
y = 5(0.2)² - 2(0.2) + 4

#

,calc 5(0.2)^2 - 2(0.2) + 4

obsidian monolithBOT
#

Result:

3.8
patent beacon
#

Not quite 4

graceful egret
#

oh

#

i was using the +4 at the end

#

not sure why i did that lol

patent beacon
#

So f(x) gives the y-coordinate, plug into that when you want the y-coord.

f'(x) gives the slope

graceful egret
#

alright i think i got it

#

honestly unless you pretend your afk im just going to keep asking for help

patent beacon
#

Go for it

#

I'll pretend I'm afk if I'm bored

graceful egret
#

haha alright😆

patent beacon
#

Jk you're in the fun early part of math

graceful egret
#

honestly i only have a few more concepts i dont know how to do

#

for this test anyway

patent beacon
#

Ask away I hope I can help

graceful egret
#

ok so i have the answers

#

but id like to go through the process

patent beacon
#

First question is legit asking for w'(t)

#

Which you seem to have no problem getting

graceful egret
#

so derivative of that whole thing

#

oh yea ok

#

i got that

patent beacon
#

b) is asking for W(11)

#

Which doesn't require calculus

graceful egret
#

so when dealing with derivatives do i always omit a coefficient standing alone?

#

just plug in 11 there.. got it

patent beacon
#

A constant gets eliminated by a derivative yeah

graceful egret
#

ok

#

for c i kind of remember

#

but need a refresher

patent beacon
#

c) Is asking for W'(11)

graceful egret
#

oh

patent beacon
#

Which is to say, plug 11 into the derivative

graceful egret
#

wati

patent beacon
#

"Rate of change" means derivative

graceful egret
#

ok

#

could you show me what plugging 11 into the derivative would look like>

#

im not quite sure i understand

#

dang guess you got bored 😆

patent beacon
#

,calc 1.98 - 0.0164(11) + 0.001044(11)^2

obsidian monolithBOT
#

Result:

1.925924
patent beacon
#

That is plugging 11 into the derivative

graceful egret
#

ahh ok

patent beacon
#

Putting t = 11 into the derivative function

graceful egret
#

plugging 11 into equation a

#

that was the deriviative

#

got it

#

ok i got another question, someone else tried to help me with this but i didn't understand them

patent beacon
#

You can factor, and cancel, (x - 13) from the numerator and denominator

#

The denominator is pretty easy, that's just a difference of squares. How does the numerator factor?

graceful egret
#

x^2+16x-13x-208/(x+13)(x-13)

#

this is as far as i got

patent beacon
#

Denominator is perfect. You've gotten nowhere with the numerator though

#

If you can't factor a quadratic, the quadratic formula always works

graceful egret
#

oh yea

patent beacon
#

But actually learn factoring techniques. Synthetic factoring is important

graceful egret
#

dude

#

is that the one

#

where you have the 0s at the end

#

and you bring it down or something

#

so im at -3 +-sqrt(832)/2

patent beacon
#

You can do that here! You know (x - 13) is a factor of x² + 3x - 208, so you can factor it out with synthetic

graceful egret
#

i honestly dont remember synthetic

#

and as bad as it may sound im honestly cramming for tomorrow

#

and dont have time to relearn that rn 😦

patent beacon
#

Using that to show that
x² + 3x - 208 = (x - 13)(x + 16)

graceful egret
#

ok but how can i get that with quadratic formula

#

was i on the right trackl?

patent beacon
#

x = -3/2 ± √[9 + 4(208)]/2

#

x = -3/2 ± √[841]/2

#

x = -3/2 ± 29/2

#

x = -16, 13

graceful egret
#

oh

patent beacon
#

Therefore it factors into the opposite of those,
(x + 16)(x - 13)

graceful egret
#

(-3 +-sqrt(841))/2

#

,calc sqareroot 841

obsidian monolithBOT
#

The following error occured while calculating:
Error: Undefined symbol sqareroot

graceful egret
#

=29

#

(-3+-29)/2

#

(-3 +29)/2=13

#

(-3 -29)/2=-16

#

ahhhh

#

wait

#

so now what??

#

oh yea

#

(x + 16)(x - 13)/(x+13)(x-13)

patent beacon
#

Exactly

graceful egret
#

(x + 16)/(x+13)

#

no

#

wait

#

I KNEW THAT HAHA

patent beacon
#

Dividing out x - 13 means the limit can now be evaluated

graceful egret
#

16/13?

patent beacon
#

,calc (13 + 16) / (13 + 13)

obsidian monolithBOT
#

Result:

1.1153846153846
patent beacon
#

Plug x = 13 in, since it's the limit as x → 13

graceful egret
#

ok

patent beacon
#

Which was impossible before since you get 0/0, but is easy now

graceful egret
#

so the answer is 1.1153846153846?

patent beacon
#

Yus

graceful egret
#

wow

#

ok

patent beacon
#

29/26

graceful egret
#

ah gotcha

#

i could try some problems on my own

#

but im afraid you wont be here when i need the help

#

any estimate on how long youll be on?

patent beacon
#

Ask whenever, I get to you or I don't. Other people know calc too!

graceful egret
#

debatable! 😄

#

thanks for the help

#

going to try some on my own

deft flume
#

hey

#

can the 12 cancel out in 4x-12/12

limber bone
#

no

deft flume
#

because its attached to the 4x?

limber bone
#

yes with a minus sign

#

cuz its subtracted from 4x

deft flume
#

alright

#

ty

#

im trying to solve for x= 8 + 4x-12/12

#

i need to practice more of these problems, solving these have became vague

#

i can just multiply 12 to both sides? to get rid of the denominator

mystic rune
#

@deft flume

#

$x=8+\frac{4x-12}{12}$

#

Is this your question?

sour plinth
#

pretty sure it's (4x-12)/12

mystic rune
#

?

obsidian monolithBOT
mystic rune
#

It doesn't help when the person asking the question decides not to reply to people trying to help

deft flume
#

lol sorry, i was busy solving it on my own

#

thx for trying to help but i solved it

#

on the contrary, there is another problem i want to kno hoe to approach

#

simple derivation?

mystic rune
#

shoot

deft flume
#

Find W: P = 2W + 2L

mystic rune
#

$P=2W+2L$

obsidian monolithBOT
mystic rune
#

lets rearrange this so W is on the left side

#

giving

deft flume
#

i attempted it and got w = l-p lol

mystic rune
#

$2W+2L=P$

obsidian monolithBOT
mystic rune
#

subtracting 2L from both sides gives

#

$2W=P-2L$

obsidian monolithBOT
mystic rune
#

and finally

#

dividing both sides by 2 gives

#

$W=\frac{P-2L}{2}$

obsidian monolithBOT
deft flume
#

ohhh i knew i questioned that part

#

idk why i keep thinking that i can easily divide the denominator by the attached variable

#

like the 2L

#

which i ended up gettinf W = L-P

mystic rune
#

Do you mean for example $\frac{2L}{2}=L$

obsidian monolithBOT
mystic rune
#

?

deft flume
#

yea

mystic rune
#

well

#

lets look at this example to see why it wouldnt work

deft flume
#

ohhh

mystic rune
#

first of all

#

the both is true

#

however

deft flume
#

now i dogured it out

mystic rune
#

above*

#

however

deft flume
#

figured

mystic rune
#

$\frac{P-2L}{2}$

obsidian monolithBOT
mystic rune
#

lets look at this closer.

#

I can rewrite this in another way

deft flume
#

i would be dividing the P as well

mystic rune
#

$\frac{P}{2}-\frac{2L}{2}$

obsidian monolithBOT
mystic rune
#

exactly

#

so I can say

deft flume
#

yeah, I've created the habit to focus on the variable that looks like an easy cancellation

mystic rune
#

$W=\frac{P}{2}-L$

obsidian monolithBOT
mystic rune
#

If I really want to

#

but

#

its 6 and half a dozen

#

🤷

#

I would leave it as it was originally

deft flume
#

yea

#

thanks a lot

mystic rune
#

np

deft flume
#

i should review some algebra again lmao

mystic rune
#

yeah definitely, you wont remember this unless you practise it

#

it may be fresh in your head right now

#

give it a day you will be making the same mistakes

#

go and solve 15 questions on this 😃

deft flume
#

I would say that I haven't efficiently learned a few algebra topics properly due to terrible teachers

mystic rune
#

Only you are responsibile for your own learning

#

teachers are there to help you

deft flume
#

true

mystic rune
#

if you dont understand something

#

dont blame the world

#

take the effort to go and study it yourself

#

as you have done by coming on this discord

#

good step