## anonymous one year ago if, for all x, f'(x)=(x-2)^4(x-1)^3 , it follows that the function f has

1. myininaya

so you need help integrating then $\int\limits (x-2)^2(x-1)^3 dx$ Let u=x-1 then du=dx if u=x-1 subtracting one on both sides gives: u-1=x-2 so you have $\int\limits_{}^{}(u-1)^2u^3 du$ this is just to make the multiplication less harsh we still have to multiply

2. myininaya

so multiply out (u-1)^2 then multiply each one of those terms by u^3 then integrate term by term

3. myininaya

oh you aren't asking to find f

4. myininaya

5. anonymous

i just started learning calculus and came across this problem on a practice test could u tell me specifically what this is called so i can search it up and learn it

6. myininaya

what is the question exactly though ?

7. anonymous

i was asking for the relative minimum

8. anonymous

i just dont kno how to approach this

9. myininaya

well you can find the critical numbers of f by setting f'=0 and also finding where f' dne (we don't have to worry about this case here because f' is a polynomial) f'=0 means we need to solve (x-2)^2(x-1)^3=0 again these x's that satisfy (x-2)^2(x-1)^3=0 are critical numbers

10. myininaya

the critical numbers is where you can possibly have relative mins or maxes

11. anonymous

oh so i just set x-1=0

12. myininaya

and x-2=0

13. myininaya

so you have critical numbers x=1 or x=2 now it is possible to have a rel max or min or neither at the critical numbers

14. anonymous

oh kk and 1 and 2 are the critical number but the relative minimum is 1 since that is the lowest

15. myininaya

I would rather not find f'' we can use f' and draw a number line in test intervals around these numbers to see if f is decreasing or increasing

16. myininaya

|dw:1439324950653:dw|

17. myininaya

well we don't know that

18. myininaya

i mean yes 1 is less than 2 but that doesn't mean f(1) is a relative min

19. myininaya

|dw:1439325010596:dw|

20. myininaya

I chose a random number from each of the intervals

21. myininaya

that were divided by the critical numbers x=1 and x=2

22. myininaya

can you find f'(0) and f'(1/2) and f'(4)?

23. myininaya

you actually only need the sign not the numerical value

24. myininaya

for example f'(0)=(-2)^4(-1)^3 all we need to know is that f'(0) is negative which means f is decreasing before we get to x=1

25. anonymous

f'(0)=- f'(.5)=- f'(4)=+

26. myininaya

|dw:1439325170086:dw|

27. myininaya

now we want to know what is happening between 1 and 2 so we need to find the sign of f'(1/2)

28. myininaya

oh let me check didn't see you post that sorry

29. myininaya

|dw:1439325249002:dw|

30. myininaya

so we have no relative max

31. myininaya

we only have a relative min

32. myininaya

at that occurs at x=2 but the rel min value is given by f(2)

33. myininaya

f(1) was neither a rel max or a rel min because the function did not have a switch from decreasing to increasing or the other way around

34. anonymous

so rel min is at x=2?

35. myininaya

yep

36. anonymous

thats also the absolute min right?

37. myininaya

|dw:1439325403944:dw| because on this interval we can see the lowest value of is f(2)

38. anonymous

oh lol we made a mistake

39. myininaya

what is the mistake

40. anonymous

its 1.5

41. myininaya

lol yes

42. anonymous

so 1 is minimum

43. myininaya

that was dumb sorry

44. anonymous

no its okay it helped me learn thnks

45. myininaya

|dw:1439325632222:dw|

46. myininaya

so we have to fix somethings

47. anonymous

i unerstand it now its k

48. myininaya

|dw:1439325670045:dw|

49. myininaya

so the graph looks something like this and no I don't know if f(2) is 0 this is just a very very rough drawing

50. myininaya

so yes you are right f(1) is rel min

51. myininaya

and f(2) is neither rel min or max

52. myininaya

and from the graph it does appear f(1) is also a abs min

53. myininaya

you were asking for examples or a site for which this is called: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx

54. anonymous

appreciate it

55. myininaya

np