## 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

1. .Sam. Group Title

$p(x)=(x+1)^2(2x-x^2)$ 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

2. Christos Group Title

its -1 0 and 2

3. Christos Group Title

Oh so thats the first step

4. .Sam. Group Title

Yup |dw:1369655367345:dw|

5. .Sam. Group Title

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

6. Christos Group Title

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

7. .Sam. Group Title

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

8. Christos Group Title

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

9. .Sam. Group Title

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

10. Christos Group Title

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

11. Christos Group Title

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

12. .Sam. Group Title

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

13. Christos Group Title

oh ok lets find the ys

14. Christos Group Title

y = 0 ?

15. Christos Group Title

I subtitute with 0 or the roots?

16. .Sam. Group Title

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

17. .Sam. Group Title

Then that's the coordinates of max and minimum points

18. Christos Group Title

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

19. .Sam. Group Title

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

20. Christos Group Title

ok I see

21. .Sam. Group Title

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

22. Christos Group Title

ok

23. Christos Group Title

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

24. .Sam. Group Title

Yup |dw:1369657703814:dw|

25. .Sam. Group Title

Then the coefficient of highest degree is negative so the graph curves down|dw:1369657774047:dw|

26. .Sam. Group Title

And yeah, zeros are -1 0 and 2, why am I plotting 1 for.

27. Christos Group Title

hmm are you plotting it to get an estimation of the infliction point?

28. .Sam. Group Title

Yup

29. .Sam. Group Title

or else you won't know where it changes it's direction

30. Christos Group Title

I see , is there anything else left for this graph/exersise?

31. .Sam. Group Title

That's it

32. Christos Group Title

Thanks :) I will try to solve some on my own now, if I have any problem can I tell you?

33. .Sam. Group Title

I'll try :)

34. Christos Group Title

ok