Question

Graphs and functions/derivatives,

Hey can you help me with this? http://screencast.com/t/MVj3e9Ok2Yg0

Answer 1

\[p(x)=(x+1)^2(2x-x^2)\]
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

Answer 2

its -1 0 and 2

Answer 3

Oh so thats the first step

Answer 4

Yup

Answer 5

Now, differentiate p(x), set it equals to zero to find the maximum and minimum points

Answer 6

its -4x^3+6x+2 if I am not mistaken we take x=0 and x=+-sqrt(3)/sqrt(2)

Answer 7

Yeah, set -4x^3+6x+2 equals zero and solve for x

Answer 8

yea I solved for x and I got 3 roots x=0 and x=+-sqrt(3)/sqrt(2) are my calculations correct??

Answer 9

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

Answer 10

hmm how did you get from line 2 to line 3?

Answer 11

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

Answer 12

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 \((2x^3-3x-1)\)

Answer 13

oh ok lets find the ys

Answer 14

y = 0 ?

Answer 15

I subtitute with 0 or the roots?

Answer 16

Substitute these \(x=\frac{1}{2} \left(1-\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\) into \(y=(x+1)^2(2x-x^2)\)

Answer 17

Then that's the coordinates of max and minimum points

Answer 18

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

Answer 19

Yeah It's abit long
For \(x=\frac{1}{2} \left(1-\sqrt{3}\right)\) or -0.3660,
\[y=(x+1)^2(2x-x^2)\]
\[y=(\frac{1-\sqrt3}{2} +1)^2(2(\frac{1-\sqrt3}{2})-(\frac{1-\sqrt3}{2})^2)\]
\[y=\frac{9-6\sqrt{3}}{4}~~or-0.348\]

Answer 20

ok I see

Answer 21

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

Answer 22

ok

Answer 23

Now that we know those coordinates how can we use them?

Answer 24

Yup