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

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.Sam.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
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
Oh so thats the first step
 10 months ago

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

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

ChristosBest 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)
 10 months ago

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

ChristosBest 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??
 10 months ago

.Sam.Best 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
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
hmm how did you get from line 2 to line 3?
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
http://screencast.com/t/yNU8TQxrMkn this here
 10 months ago

.Sam.Best 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 \((2x^33x1)\)
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
oh ok lets find the ys
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
I subtitute with 0 or the roots?
 10 months ago

.Sam.Best 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 \(y=(x+1)^2(2xx^2)\)
 10 months ago

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

ChristosBest 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
 10 months ago

.Sam.Best 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, \[y=(x+1)^2(2xx^2)\] \[y=(\frac{1\sqrt3}{2} +1)^2(2(\frac{1\sqrt3}{2})(\frac{1\sqrt3}{2})^2)\] \[y=\frac{96\sqrt{3}}{4}~~or0.348\]
 10 months ago

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

ChristosBest ResponseYou've already chosen the best response.0
Now that we know those coordinates how can we use them?
 10 months ago

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

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

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

ChristosBest ResponseYou've already chosen the best response.0
hmm are you plotting it to get an estimation of the infliction point?
 10 months ago

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

ChristosBest ResponseYou've already chosen the best response.0
I see , is there anything else left for this graph/exersise?
 10 months ago

ChristosBest 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?
 10 months ago
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