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Graphs and functions/derivatives, Hey can you help me with this?

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\[p(x)=(x+1)^2(2x-x^2)\] From there, you can tell that (x+1)=0, x=-1 is a zero of p(x) Then factor this \((2x-x^2)\) and set the brackets equals zero and find the other 2 zeros
its -1 0 and 2
Oh so thats the first step

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Yup |dw:1369655367345:dw|
Now, differentiate p(x), set it equals to zero to find the maximum and minimum points
its -4x^3+6x+2 if I am not mistaken we take x=0 and x=+-sqrt(3)/sqrt(2)
Yeah, set -4x^3+6x+2 equals zero and solve for x
yea I solved for x and I got 3 roots x=0 and x=+-sqrt(3)/sqrt(2) are my calculations correct??
Hmm, \[-4x^3+6x+2 \\ \\ -2(2x^3-3x-1) \\ \\ -2(x+1)(2x^2-2x-1)\] Factoring \(2x^2-2x-1\) 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 \[y=(x+1)^2(2x-x^2)\] Find the y's
hmm how did you get from line 2 to line 3? this here
I used trial an error with calculator, when \[p(-1)=-2(2(-1)^3-3(-1)-1)=0\] Then (x+1) is a factor of p(x), then use (x+1) divide \((2x^3-3x-1)\)
1 Attachment
oh ok lets find the ys
y = 0 ?
I subtitute with 0 or the roots?
Substitute these \(x=\frac{1}{2} \left(1-\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\) into \(y=(x+1)^2(2x-x^2)\)
Then that's the coordinates of max and minimum points
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
Yeah It's abit long For \(x=\frac{1}{2} \left(1-\sqrt{3}\right)\) or -0.3660, \[y=(x+1)^2(2x-x^2)\] \[y=(\frac{1-\sqrt3}{2} +1)^2(2(\frac{1-\sqrt3}{2})-(\frac{1-\sqrt3}{2})^2)\] \[y=\frac{9-6\sqrt{3}}{4}~~or-0.348\]
ok I see
(-0.3660,-0.3481) Then for the other coordinate is (1.366,4.848)
Now that we know those coordinates how can we use them?
Yup |dw:1369657703814:dw|
Then the coefficient of highest degree is negative so the graph curves down|dw:1369657774047:dw|
And yeah, zeros are -1 0 and 2, why am I plotting 1 for.
hmm are you plotting it to get an estimation of the infliction point?
or else you won't know where it changes it's direction
I see , is there anything else left for this graph/exersise?
That's it
Thanks :) I will try to solve some on my own now, if I have any problem can I tell you?
I'll try :)

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