lilfayfay 3 years ago Let f be the function defined by f(x)=3x^5 - 5x^3 +2 (a) On what intervals is f increasing? (b) On what intervals is the graph of f concave upward? (c) Write the equation of each horizontal tangent line to the graph of f.

1. Goten77

1st derivative test?

2. lilfayfay

do first derivative test?

3. lilfayfay

@Goten77 ok for first derivative i got 15x^4 - 15x^2

4. Goten77

k thats right now factor it

5. Goten77

|dw:1358134355078:dw|

6. Goten77

set each = to 0 and get critical points |dw:1358134445508:dw| but also find where x never exist.. but it always does so ignore that part

7. Goten77

now set up the critical points to test for when its positive and negative and you use the 1st derivative when plugging in theese values|dw:1358134521331:dw| u just use values in between the numbers

8. Goten77

so thats part a when its increasing/decreasing.... also part c is just the graphs of y = 0 , -1, +1 since those are the 3 horizontal tangents

9. Goten77

part b is 2nd derivative test

10. Goten77

|dw:1358134818974:dw|

11. lilfayfay

for part a.) is it (-infinity,-1) & (1,infinity)

12. Goten77

yes

13. lilfayfay

ok for part b.) i got x=0, and $\pm \frac{ \sqrt{2} }{ 2 }$

14. Goten77

|dw:1358135080380:dw|

15. Goten77

|dw:1358135162509:dw| hmm you number works to...

16. lilfayfay

then i do the SDT? (Second Derivative Test)?

17. Goten77

well there the same i meant XD

18. Goten77

yes now you test those numbers between/around them

19. lilfayfay

ok, how do i do part c.)? o:

20. Goten77

i told u part c

21. Goten77

so um heres part b|dw:1358135773604:dw|

22. Goten77

hmm the squareroot number is very small....

23. lilfayfay

for b.) did u get $(\frac{ \sqrt{2} }{ 2 }, \infty)$

24. Goten77

oh i keep on thinking about point of inflection <.<

25. lilfayfay

|dw:1358128938034:dw| o.O

26. lilfayfay

is that drawing seem right? :\

27. Goten77

|dw:1358136164439:dw|

28. Goten77

yes your right... its only concave up on that last interval

29. lilfayfay

:"D

30. lilfayfay

so how do i do part c.) again sorry..

31. lilfayfay

wait so my part b.)was wrong?

32. Goten77

so heres what the graph looks like |dw:1358136607164:dw|

33. Goten77

aah i meant to say for part (c)

34. Goten77

k everything is right now.... i just got part letters mixed up

35. lilfayfay

so how do ido part c.)

36. Goten77

......u kno what i meant... that entre paragraph just now is part C

37. Goten77

oh i said it kinda wrong earlier i meant part Cis when f(x)= -1,0,1 meaning plug those 3 values into the original function and thats when the graph has a tangent lind so then you get y= 4 y= 2 y= 0 and thats when the graph has the horizontal tangents the x values of -1,0,1 came from the 1st derivative test since those are the critical points meaning where the derivative = 0

38. lilfayfay

ah ok ty so for part c.) i got y=-2x+2

39. Goten77

no i gave u part 2

40. Goten77

part c*

41. lilfayfay

yeah it says write the equation

42. Goten77

ill explain it again

43. Goten77

k so the 1st derivative gives us critical points meaning where the slope = 0 also meaning at those points on the original graph is where it has a horizontal tangent..... so taht means the horizontal tangents are in the form of y = # and thats it so our critical points were -1 0 and 1 plug those into the original equasion and get y= 4 y= 2 y= 0 and thats it .. those are your 3 horizontal tangents

44. lilfayfay

ohso my answer waswrong x-x

45. lilfayfay

okty :P

46. lilfayfay

@Goten77 canu help me with another 2 more pls? ^^

47. Goten77

sure