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ParthKohliBest ResponseYou've already chosen the best response.1
As they always say: find \(f'(x)\) and equate that to \(0\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
@chochko What would \(f'(x)\) be here?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Well that's correct  now equate it to zero.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Solve this equation:\[\rm 4x^3 + 12x^2 = 0\]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Yeah, I meant 4x^3 + 12x^2 = 0
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
sry my computer is really slow. Would you factor out a 4?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
Yu.. 4x^2 ( x3) = 0 4x^2 = 0 x = 0 x3 = 0 x = 3
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Mine too, and you can factor 4x^2 out.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Yep, and what Yahoo! said are the critical points. ;)
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
thank you but im not done . that is just one part..can you help yahoo or ParthKohli?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
the next part is to differentiate it again :P to determine its nature
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
You guys must be in the same school.
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
using the second derivitive test which is 12x^2+24x=0 set to 0 to determine all relative extrema, indicating the x and Y values and wheteher is a max or a min. Help please
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
lol @ParthKohli: the next step is common sense. @chochko: with the differentiataed function sub the \(x\) that you found in the first step if its a negative then its concave down => max. if its concave up then => min
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Sorry, was on another question, and yeah  that's the way I was taught too.
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
so in 12x^2+24x Sub the x for the critical numbers and the result is the relative extrema? is that the Max and Min as well?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
max or min concave up => min concave down => max
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
so which part is the relative extrema? i really suck at math btw?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
well i googled; relative extrema and max/ min extrema is the same thing
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
f''(0)=0 min f''(3)=36 Max Is this correct?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
i dont know; i dont have a calculator with me but if its you have to check if it has a horizontal point of inflexion
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
Ok, so the intervals are the critical points right? where are they increasing and decreasing?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
critical points are the stationary points on the curve since \(f(x)=0\) you differentiate it twice to determine if its increasing or decreasing increasing \(f''(x) >0\)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
i think i made a mistake; increasing \(f'(x)>0\)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
when the function is decreasing \(f'(x)<0\)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
i mean you differetiate it once; to see if its increasing or decreasing sorry; im sick
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
so using the f'(x)=0 and plug in 0 and 3 for x to find where f(x) is increasing or decreasing?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
oh no no im getting tired @ParthKohli: may like to explain :)
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
@ParthKohli are you able to help me finish this one
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
I think I should explain; since Parth is hiding :P \(f'(x) = 0\) you are finding the \(x\) values that is the STATIONARY point. \(f'(x) >0\) is the \(x\) values were the function is INCREASING \(f'(x)<0\) is where the function is DECREASING. The second derivative is the deirivative of the derivative. The curve \(y=f(x)\) is concave up when \(f''(x)>0\) and concve down when \(f''(x)<0\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Didn't notice the tag right there. :P
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
Increasing on ( inf, 0) (0,3) then decreasing (3, inf) idk if its correct?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Well your \(\rm f'(x)\) function is \(\rm 4x^2 +12x \) and the function is stationary at x = 3,0.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
What is the question again?
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
State the intervas for which f(x) is increasing or decreasing So i tested the values 1, 1, 4 And got F'(1)=16 inc f'(2)=8 inc f'(4)=64 dec. idk if its correct?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
So it starts decreasing after 3.
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
(inf, 0) (0,3) increasing then (3, inf) decreasing Are the intervals right?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Let's test f'(5). f'(x) = 4x^2 + 12x f'(5) = 4(5)^2 + 12(5) = 4(25)  60 = 100  60 = 160
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
It's decreasing at f(5).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
The best way is to ask Wolfram for the graph. :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Enter your function and see its graph.
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
ok can you help me find the possible inflection points
 one year ago

chochkoBest ResponseYou've already chosen the best response.0
@ParthKohli can you help just a lil bit left
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Damn... I've disabled the tag notifs.
 one year ago
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