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

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.Sam.
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Oh so thats the first step

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

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

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

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

.Sam.
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok lets find the ys

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

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

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

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

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

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

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

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

Christos
 3 years 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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