anonymous
  • anonymous
where is everyone?!? help please! =( find the extrema: 4x^3-13x^2+4x+1 max and min: [1,4] [0,2]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
so i found the derivative: 12x^2 -26x +4
anonymous
  • anonymous
correct?
anonymous
  • anonymous
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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anonymous
  • anonymous
yup.
anonymous
  • anonymous
do you know hot to factor: 12x^2 -26x +4
anonymous
  • anonymous
I'll try and see if I can help but can't promise so yeah you have the derivative right
anonymous
  • anonymous
not on calculator though... i just need to know just in case.
anonymous
  • anonymous
i must factor first in order to = to 0
anonymous
  • anonymous
so to find max and min you find the derivative set it = to 0 and solve?
anonymous
  • anonymous
yes i have to find the critical points.. then evaluate the point to original function
anonymous
  • anonymous
then evaluate with the endpoints. then i'll find max and min
anonymous
  • anonymous
so you have 1/6 and 2 for the critical points right?
anonymous
  • anonymous
@campbell_st @precal @heradog the more help the better! :( i really need to figure this problem out.
anonymous
  • anonymous
well.... i don't know how to factor this exactly: 12x^2 -26x +4
anonymous
  • anonymous
since derivative is right... then how do i factor it out?
anonymous
  • anonymous
@dan815
precal
  • precal
|dw:1363327156303:dw|
anonymous
  • anonymous
yes @precal that one is right
precal
  • precal
|dw:1363327198254:dw|
anonymous
  • anonymous
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)
anonymous
  • anonymous
hm... i did that also... but then i have to find x . critical points.
precal
  • precal
|dw:1363327232310:dw|
precal
  • precal
|dw:1363327276945:dw|
precal
  • precal
are you given those intervals? [1,4] and [0,2]
campbell_st
  • campbell_st
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
precal
  • precal
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
anonymous
  • anonymous
kindly, can we stick with these steps.... {Differentiate} {Set f '(x) = 0} {Factor} {solve to get the critical numbers}
anonymous
  • anonymous
im already used to them...
precal
  • precal
|dw:1363327423329:dw|
anonymous
  • anonymous
@precal what did you do after 2(6x^2-13x+2) ? i dont understand how you got the x^2....etc after...
precal
  • precal
|dw:1363327470127:dw|
precal
  • precal
|dw:1363327620907:dw|it is an old jedi trick called slide and divide, makes factoring easier to do
campbell_st
  • campbell_st
the attached image may help
1 Attachment
precal
  • precal
max at x=1/6 and min at x=2
anonymous
  • anonymous
|dw:1363327660835:dw|
precal
  • precal
|dw:1363327725416:dw|
anonymous
  • anonymous
what i meant t say how? i know you said a trick way but i don't know how... :/
campbell_st
  • campbell_st
the 2nd derivative test is used for determining if the point is a max or min... easier than 1st der and change of sign...
precal
  • precal
true but if this is for a calculus ab class, they are taught to do 1st d/dx to determine max and min
anonymous
  • anonymous
stilll dont understand .. :(
precal
  • precal
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......
campbell_st
  • campbell_st
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
precal
  • precal
like I said earlier, use whatever works for you. This problem is beyond factoring......
precal
  • precal
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
anonymous
  • anonymous
yes i used that too. i mea im fine with everything except factoring now.
anonymous
  • anonymous
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.
precal
  • precal
well I like to use slide and divide
anonymous
  • anonymous
i never used that technique before
precal
  • precal
http://mrsgalgebra.pbworks.com/w/page/12019090/Slide%20and%20Divide%20Method%20for%20Factoring%20Polynomials
precal
  • precal
look it up, you will find many videos on it on youtube. good luck. factoring gets easier over time.
anonymous
  • anonymous
thanks :)
precal
  • precal
yw
anonymous
  • anonymous
@precal
anonymous
  • anonymous
i put the answer: [1,4] max=65 min= -11 and for [0,2] max= 196 min= -11 what's wrong with the answer?
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
somebody help pls!
precal
  • precal
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
precal
  • precal
gotta go will help you tomorrow, it is very late for me at the moment (past 2 am)

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