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

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.Sam. Group TitleBest 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
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
its 1 0 and 2
 one year ago

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

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

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

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

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

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

.Sam. Group TitleBest ResponseYou've already chosen the best response.3
Hmm, \[4x^3+6x+2 \\ \\ 2(2x^33x1) \\ \\ 2(x+1)(2x^22x1)\] 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
 one year ago

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

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

.Sam. Group TitleBest ResponseYou've already chosen the best response.3
I 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 \((2x^33x1)\)
 one year ago

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

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

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

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

Christos Group TitleBest ResponseYou've already chosen the best response.0
that 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
 one year ago

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

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

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

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

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

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

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

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

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

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