At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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its -1 0 and 2

Oh so thats the first step

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 how did you get from line 2 to line 3?

http://screencast.com/t/yNU8TQxrMkn this here

oh ok lets find the ys

y = 0 ?

I subtitute with 0 or the roots?

Then that's the coordinates of max and minimum points

ok I see

(-0.3660,-0.3481)
Then for the other coordinate is
(1.366,4.848)

ok

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?

Yup

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 :)

ok