Lmao..

okay ummm... rules... I'll leave a question, then you answer it & leave a question for me... first to get three wrong loses... and you have to be able to do the question yourself... must show working...

g'luck BITCH!!

First Question.

Differentiate the following equation with respect to x.

y = (3x - 5)^6 x (-2x^2 + 7)^4

This should be good. :)

damn. Rapworlds Geeks. lol g'luck you 2.

y = (3x - 5)^6 x (-2x^2 + 7)^4
y= 6(3x - 5)^5 x 4(-2x^2 + 7)^3


I hope I'm right or else Imma kill myself.


Anyway... What are the minimum and maximum values of the equation:

y = 2x^3 - 15x^2 + 36x - 20

*wrong buzzer sounds*

bwahaha... product rule + chain rule... it's

y = (3x - 5)^6 x (-2x^2 + 7)^4
y' = (6)(3)(3x - 5)^5(-2x^2 + 7)^4 + (3x - 5)^6(4)(-4x)(-2x^2 + 7)^3

----

your question:

y = 2x^3 - 15x^2 + 36x - 20

y' = 6x^2 - 30x +36

minima/maxima occur when the gradient function = 0

0 = 6x^2 - 30x +36
0 = (2x - 6)(3x - 6)

Null factor law: 2x - 6 = 0 or 3x - 6 = 0

stationary points occur when x = 3 & 2

y co-cordinates:

y(3) = 54 - 135 + 108 - 20 = 7
y(2) = 16 - 60 + 72 - 20 = 8

to determine the nature of each stationary point:

y" = 12x - 30

y"(3) = 36 - 30 = 6 = positive

therefore there is a local minimum at (3,7)

y"(2) = 24 - 30 = -6 = negative

therefore there is a local maximum at (2,8)

I Just Gained Like 10 Percent More To My IQ By Looking At That...

..Wow Nuke, Pissy sonned you.. Lol. That's some shit though, I don't know anything about that... It seems interesting though... I didn't make it through school long enough to learn it, hehe...

Ok second question:

Use calculus to find the EXACT area of the region bounded by the graphs with equations y = cos(x) and y = cos(2x) for 0 < x < 2pi (those should be less than OR equal to signs, but I dunno how to type them, so just imagine them lol)

nuke got merked. :(

Ok second question:

Use calculus to find the EXACT area of the region bounded by the graphs with equations y = cos(x) and y = cos(2x) for 0 < x < 2pi (those should be less than OR equal to signs, but I dunno how to type them, so just imagine them lol)

Haven't done anything with pi and graphs, yet... another question, please. And :mad: at getting the first one wrong... I knew it was wrong but just didn't know what to do... TRY ME NOW BITCH!

what is the square root of my dick? :mad:

^2................. :D <--the fuck @ that face not workin

In that advanced math thread you knew how to integrate to find area didnt you ?? Or do you mean you cant do the trig part of it?

OK... ummm find the exact area of the region bounded by the y-axis, the graph of y = x^2 - 4 and the line x = 3

In that advanced math thread you knew how to integrate to find area didnt you ?? Or do you mean you cant do the trig part of it?

OK... ummm find the exact area of the region bounded by the y-axis, the graph of y = x^2 - 4 and the line x = 3

Smh... why a graph question?!

*goes to find some paper*



-



y = x^2 - 4
x = 3

S3~0 (x^2 - 4)dx
= [1/3x^3 - 4x]3~0

So

[1/3.3^3 - 4.3] - [1/3.0^3 - 4.0]
[3 - 12] - [0]
= -9


I havent done a question like this since last year... I think its wrong. I think I've missed a step out.






Anyway... erm... your question

f(x) = 5x^3 + 3x^2 + 7x - 9
g(x) = 2x^s + 6x^2 - 8x - 11

Show that S2~1 {f(x) - g(x)dx = S2~1 f(x)dx - S2~1 g(x)dx





Dunno if it makes sense... S2~1 = Integrate with limits of 2 and 1... S is like the sign.. that elongated "S".. w/e.. smh

Oh god.. I thought I was smart. Fuck maths..

word^^^

More or less right nuke... mostly right... you just forgot to switch the terminals/take the absolute value bc the area was below the x axis... cant have a neg area... but yea it's 9 square units... you pass that one...

-----

f(x) = 5x^3 + 3x^2 + 7x - 9
g(x) = 2x^s + 6x^2 - 8x - 11

Show that S2~1 {f(x) - g(x)dx = S2~1 f(x)dx - S2~1 g(x)dx

------


ermm... because Sa~b{c(x) - d(x)}dx = Sa~b(c(x))dx - Sa~b(d(x))dx ... it's just a rule of integration...

but anyway...

I assume that s is supposed to be a 3...

S2~1{f(x) - g(x)}dx = S2~1{5x^3 + 3x^2 + 7x - 9 - 2x^3 - 6x^2 + 8x + 11}dx

= S2~1(3x^3 - 3x^2 +15x + 2)dx

= [.25x^4 - x^3 +7.5x^2 +2x]2~1

= [4 - 8 + 15 +4] - [.25 - 1 +7.5 +2]

= 15 - 8.75

= 6.25

..........

S2~1 f(x)dx = S2~1(5x^3 + 3x^2 + 7x - 9)dx

= [1.25x^4 + x^3 + 3.5x^2 - 9x]2~1

= [20 + 8 + 14 - 18] - [1.25 + 1 + 3.5 - 9]

= 24 - -3.25

= 27.25


finsh later, gotta go

x = who gives a fuck

.....take the anti-derivative of my nuts...multplied by the inverse proportion of the area of nut on each of ur mother's eyes...then graph it using the cossintan over a function of X......





I Expect answers soon.......:(

A forum for homework? And Math Battles? Oh my God. This is a new low. It's pretty hard to get any lower than Calculus Square Offs. This really addresses the heart and culture of rap and hip hop. I gotta stay away from this site for a while after seeing this. I can't let my daughter's mother see me looking at this stuff. Gotta go :mad:

Your Worst Enemy= No Deal
Yin Yang Twins= Multi Platinum

.....^^^low blow......lmao.....




*OWNED*

I Just Gained Like 10 Percent More To My IQ By Looking At That...


I lost around 10% from looking at that...


So now my IQ hangs somewhere around the -40 mark... :(

lmao^

I'll gladly take this as a forefeit win :D. He was doing it wrong anyway. :rolleyes: