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

- anonymous

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

- jamiebookeater

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- anonymous

so i found the derivative: 12x^2 -26x +4

- anonymous

correct?

- 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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## More answers

- anonymous

yup.

- anonymous

do you know hot to factor: 12x^2 -26x +4

- anonymous

I'll try and see if I can help but can't promise
so yeah you have the derivative right

- anonymous

not on calculator though... i just need to know just in case.

- anonymous

i must factor first in order to = to 0

- anonymous

so to find max and min you find the derivative set it = to 0 and solve?

- anonymous

yes i have to find the critical points.. then evaluate the point to original function

- anonymous

then evaluate with the endpoints. then i'll find max and min

- anonymous

so you have 1/6 and 2 for the critical points right?

- anonymous

@campbell_st @precal @heradog the more help the better! :( i really need to figure this problem out.

- anonymous

well.... i don't know how to factor this exactly: 12x^2 -26x +4

- anonymous

since derivative is right... then how do i factor it out?

- anonymous

@dan815

- precal

|dw:1363327156303:dw|

- anonymous

yes @precal that one is right

- precal

|dw:1363327198254:dw|

- 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

hm... i did that also... but then i have to find x . critical points.

- precal

|dw:1363327232310:dw|

- precal

|dw:1363327276945:dw|

- precal

are you given those intervals? [1,4] and [0,2]

- 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

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

kindly, can we stick with these steps....
{Differentiate}
{Set f '(x) = 0}
{Factor}
{solve to get the critical numbers}

- anonymous

im already used to them...

- precal

|dw:1363327423329:dw|

- anonymous

@precal what did you do after 2(6x^2-13x+2) ?
i dont understand how you got the x^2....etc after...

- precal

|dw:1363327470127:dw|

- precal

|dw:1363327620907:dw|it is an old jedi trick called slide and divide, makes factoring easier to do

- campbell_st

the attached image may help

##### 1 Attachment

- precal

max at x=1/6 and min at x=2

- anonymous

|dw:1363327660835:dw|

- precal

|dw:1363327725416:dw|

- anonymous

what i meant t say how? i know you said a trick way but i don't know how... :/

- 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

true but if this is for a calculus ab class, they are taught to do 1st d/dx to determine max and min

- anonymous

stilll dont understand .. :(

- 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

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

like I said earlier, use whatever works for you. This problem is beyond factoring......

- 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

yes i used that too. i mea im fine with everything except factoring now.

- 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

well I like to use slide and divide

- anonymous

i never used that technique before

- precal

http://mrsgalgebra.pbworks.com/w/page/12019090/Slide%20and%20Divide%20Method%20for%20Factoring%20Polynomials

- precal

look it up, you will find many videos on it on youtube. good luck. factoring gets easier over time.

- anonymous

thanks :)

- precal

yw

- anonymous

@precal

- 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

@dan815

- anonymous

somebody help pls!

- 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

gotta go will help you tomorrow, it is very late for me at the moment (past 2 am)

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