## Christos 2 years ago Graphs and functions/derivatives, Hey can you help me with this? http://screencast.com/t/MVj3e9Ok2Yg0

1. .Sam.

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

its -1 0 and 2

3. Christos

Oh so thats the first step

4. .Sam.

Yup |dw:1369655367345:dw|

5. .Sam.

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

6. Christos

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

7. .Sam.

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

8. Christos

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

9. .Sam.

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

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

11. Christos

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

12. .Sam.

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

oh ok lets find the ys

14. Christos

y = 0 ?

15. Christos

I subtitute with 0 or the roots?

16. .Sam.

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.

Then that's the coordinates of max and minimum points

18. Christos

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.

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

ok I see

21. .Sam.

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

22. Christos

ok

23. Christos

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

24. .Sam.

Yup |dw:1369657703814:dw|

25. .Sam.

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

26. .Sam.

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

27. Christos

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

28. .Sam.

Yup

29. .Sam.

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

30. Christos

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

31. .Sam.

That's it

32. Christos

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

33. .Sam.

I'll try :)

34. Christos

ok