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where is everyone?!? help please! =( find the extrema: 4x^3-13x^2+4x+1 max and min: [1,4] [0,2]

Mathematics
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so i found the derivative: 12x^2 -26x +4
correct?
I could maybe figure it out for myself on my calculator but I don't know how that would help, are you in webworks?

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Other answers:

yup.
do you know hot to factor: 12x^2 -26x +4
I'll try and see if I can help but can't promise so yeah you have the derivative right
not on calculator though... i just need to know just in case.
i must factor first in order to = to 0
so to find max and min you find the derivative set it = to 0 and solve?
yes i have to find the critical points.. then evaluate the point to original function
then evaluate with the endpoints. then i'll find max and min
so you have 1/6 and 2 for the critical points right?
@campbell_st @precal @heradog the more help the better! :( i really need to figure this problem out.
well.... i don't know how to factor this exactly: 12x^2 -26x +4
since derivative is right... then how do i factor it out?
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yes @precal that one is right
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ok so to factor you pull out the recurring elements I so suck at factoring but lets try it you can multiply every element by 2 giving you 2(6x^2-13x+2)
hm... i did that also... but then i have to find x . critical points.
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are you given those intervals? [1,4] and [0,2]
the stationary points are at (x - 2)(12x -2) = 0 so the stationary points are x = 2, x = 1/6 so find the the 2nd derivative and test them for max and min then check you boundary values... an easy way it to graph the curve
why not just do a sign analysis with the first d/dx, second d/dx states concave up or concave down doesn't 1st d/dx show max and min
kindly, can we stick with these steps.... {Differentiate} {Set f '(x) = 0} {Factor} {solve to get the critical numbers}
im already used to them...
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@precal what did you do after 2(6x^2-13x+2) ? i dont understand how you got the x^2....etc after...
|dw:1363327470127:dw|
|dw:1363327620907:dw|it is an old jedi trick called slide and divide, makes factoring easier to do
the attached image may help
1 Attachment
max at x=1/6 and min at x=2
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|dw:1363327725416:dw|
what i meant t say how? i know you said a trick way but i don't know how... :/
the 2nd derivative test is used for determining if the point is a max or min... easier than 1st der and change of sign...
true but if this is for a calculus ab class, they are taught to do 1st d/dx to determine max and min
stilll dont understand .. :(
just factor using whatever technique or strategy you were taught, factoring is a basic skill you will need to for future math courses. I just used a technique I like that helps me factor faster......
ok... here is how I factor it and it works every time given \[ax^2 + bx + c = 0\] multiply a by c in your question 6 times 2 = 12 find the factors of 12 that add to -13....-12, -1 then you have (ax + factor1)(ax + factor 2) (6x -12)(6x -1) -------------------------- -------------- = 0 a 6 find the common factors in the numerator 6(x -2)(6x -1) = 0 ----------------- 6 cancel the common factors leaves (x -2)(6x -1) = 0 easy to solve
like I said earlier, use whatever works for you. This problem is beyond factoring......
kindly, can we stick with these steps.... {Differentiate} {Set f '(x) = 0} {Factor} {solve to get the critical numbers} btw I did utilize the steps above
yes i used that too. i mea im fine with everything except factoring now.
i know it was a basic thing we have to know but i forgot. so I'm trying to find an easy way to learn.
well I like to use slide and divide
i never used that technique before
http://mrsgalgebra.pbworks.com/w/page/12019090/Slide%20and%20Divide%20Method%20for%20Factoring%20Polynomials
look it up, you will find many videos on it on youtube. good luck. factoring gets easier over time.
thanks :)
yw
i put the answer: [1,4] max=65 min= -11 and for [0,2] max= 196 min= -11 what's wrong with the answer?
somebody help pls!
are you looking for absolute max or absolute min? are you looking for global or local max or min? seems like you are doing something else other than looking for critical numbers
gotta go will help you tomorrow, it is very late for me at the moment (past 2 am)

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