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 one year ago
Graphs and functions/derivatives,
Hey can you help me with this? http://screencast.com/t/MVj3e9Ok2Yg0
 one year ago
Graphs and functions/derivatives, Hey can you help me with this? http://screencast.com/t/MVj3e9Ok2Yg0

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.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3\[p(x)=(x+1)^2(2xx^2)\] From there, you can tell that (x+1)=0, x=1 is a zero of p(x) Then factor this \((2xx^2)\) and set the brackets equals zero and find the other 2 zeros

Christos
 one year ago
Best ResponseYou've already chosen the best response.0Oh so thats the first step

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Yup dw:1369655367345:dw

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Now, differentiate p(x), set it equals to zero to find the maximum and minimum points

Christos
 one year ago
Best ResponseYou've already chosen the best response.0its 4x^3+6x+2 if I am not mistaken we take x=0 and x=+sqrt(3)/sqrt(2)

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, set 4x^3+6x+2 equals zero and solve for x

Christos
 one year ago
Best ResponseYou've already chosen the best response.0yea I solved for x and I got 3 roots x=0 and x=+sqrt(3)/sqrt(2) are my calculations correct??

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Hmm, \[4x^3+6x+2 \\ \\ 2(2x^33x1) \\ \\ 2(x+1)(2x^22x1)\] Factoring \(2x^22x1\) you get \[x=\frac{1}{2} \left(1\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\] These x are the max and min points of the graph, then use these x substitute into \[y=(x+1)^2(2xx^2)\] Find the y's

Christos
 one year ago
Best ResponseYou've already chosen the best response.0hmm how did you get from line 2 to line 3?

Christos
 one year ago
Best ResponseYou've already chosen the best response.0http://screencast.com/t/yNU8TQxrMkn this here

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3I used trial an error with calculator, when \[p(1)=2(2(1)^33(1)1)=0\] Then (x+1) is a factor of p(x), then use (x+1) divide \((2x^33x1)\)

Christos
 one year ago
Best ResponseYou've already chosen the best response.0oh ok lets find the ys

Christos
 one year ago
Best ResponseYou've already chosen the best response.0I subtitute with 0 or the roots?

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Substitute these \(x=\frac{1}{2} \left(1\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\) into \(y=(x+1)^2(2xx^2)\)

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Then that's the coordinates of max and minimum points

Christos
 one year ago
Best ResponseYou've already chosen the best response.0that was really long to find wasnt it? I tried to find the first one and I found (2 + sqrt(3))/2 which seems to be wrong here

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Yeah It's abit long For \(x=\frac{1}{2} \left(1\sqrt{3}\right)\) or 0.3660, \[y=(x+1)^2(2xx^2)\] \[y=(\frac{1\sqrt3}{2} +1)^2(2(\frac{1\sqrt3}{2})(\frac{1\sqrt3}{2})^2)\] \[y=\frac{96\sqrt{3}}{4}~~or0.348\]

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3(0.3660,0.3481) Then for the other coordinate is (1.366,4.848)

Christos
 one year ago
Best ResponseYou've already chosen the best response.0Now that we know those coordinates how can we use them?

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Yup dw:1369657703814:dw

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3Then the coefficient of highest degree is negative so the graph curves downdw:1369657774047:dw

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3And yeah, zeros are 1 0 and 2, why am I plotting 1 for.

Christos
 one year ago
Best ResponseYou've already chosen the best response.0hmm are you plotting it to get an estimation of the infliction point?

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.3or else you won't know where it changes it's direction

Christos
 one year ago
Best ResponseYou've already chosen the best response.0I see , is there anything else left for this graph/exersise?

Christos
 one year ago
Best ResponseYou've already chosen the best response.0Thanks :) I will try to solve some on my own now, if I have any problem can I tell you?
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