## anonymous 3 years ago ok this is mine $f(x)=\sqrt[3]{x-1}+\sqrt[3]{3-x}$find all $$x \in \mathbb{R}$$ such that $$f(x) \in \mathbb{N}$$

1. anonymous

f(x)=2 and 1

2. anonymous

how?

3. anonymous

@nipunmalhotra93 @sauravshakya post complete solution ... i'll check it later

4. anonymous

Take it to the form a^3+b^3

5. anonymous

|dw:1347716779899:dw|

6. anonymous

Wll that be enough

7. anonymous

So, f(x) can only be either 1 or 2

8. anonymous

So the hard part is to find x

9. anonymous

Was that correct @mukushla

10. anonymous

x=2

11. anonymous

thats correct...complete it :) can u find the exact range of f(x) ?

12. anonymous

|dw:1347720322437:dw|

13. nipunmalhotra93

well, here's my solution. |dw:1347734655828:dw||dw:1347734803553:dw|

14. nipunmalhotra93

|dw:1347736319687:dw|

15. nipunmalhotra93

|dw:1347736404280:dw|

16. nipunmalhotra93

so three answers in total.... :)

17. anonymous

Fantastic work @nipunmalhotra93 :)

18. anonymous

i think the answer is only x=2

19. anonymous

|dw:1347769714174:dw|

20. anonymous

I think that also works

21. anonymous

that gives f(x)=1 ?

22. anonymous

YEP

23. anonymous

And is solved the equation

24. anonymous
25. anonymous

WHAT! so my solution was wrong

26. anonymous

because the range of that function is$[\sqrt[3]{2},2]$

27. anonymous

Cant it be zero as x---->infinity

28. anonymous

Oh I got it there is a problem with my solution

29. anonymous

So if (x)^1/2 + y^1/2 =2........i and x^1/2 - y^1/2 =4.......ii

30. anonymous

It has no solution........ though we get the value of x and y..... if we solve those equation

31. anonymous

Same thing happened to my equations

32. anonymous

right? @mukushla

33. anonymous

where that equations come from?

34. anonymous

Actually the equations I got was: (x-1)^1/3 - (x-3)^1/3 =1...........i (x-1)^2/3 +{(x-1)(x-3)}^1/3 + (x-3)^2/3=2........ii

35. anonymous

Solved those two equations

36. anonymous

Let x-1=A and x-3=B

37. anonymous

Square first equation

38. anonymous

|dw:1347770764883:dw|

39. nipunmalhotra93

@mukushla actually I think that wolfram uses the domain of $\sqrt[3]{x}$to be only non negative real nos. I have noticed that it gives complex answers when we try to plug in negative reals in the cube root function http://www.wolframalpha.com/input/?i=%5Csqrt%5B3%5D%7B-1%7D

40. anonymous

yeah thats right...have u checked ur solutions?

41. nipunmalhotra93

yup! when I plug in x=.98165, calculate cube root of 1-x and then put a minus sign with it. That works fine.

42. nipunmalhotra93

similarly, when I put x=3.0183 , I calculate x-3 and put minus sig with the cube root of x-3 to get $\sqrt[3]{3-x}$ That's the only possible way to check it I think.

43. anonymous

@nipunmalhotra93 Actually u and I got the same answers....|dw:1347775568533:dw| But And the substituted the value |dw:1347775753579:dw| in f(x) I got 1

44. anonymous

@mukushla can u PLZ CHECK it

45. anonymous

I think the range of f(x) is|dw:1347775870157:dw|

46. anonymous

thats right :)

47. anonymous

good job saura :)

48. anonymous

But I still have a question for u

49. anonymous

guys @sauravshakya @nipunmalhotra93 u cooked the problem :D

50. anonymous

yeah what ?

51. anonymous

(x-1)^1/3 - (x-3)^1/3 =1...........i (x-1)^2/3 +{(x-1)(x-3)}^1/3 + (x-3)^2/3=2........ii

52. anonymous

There I solved for x and got the value

53. anonymous

Similarly if I solved this (x)^1/2 + y^1/2 =2........i and x^1/2 - y^1/2 =4.......ii

54. anonymous

Will I get the values of x and y

55. anonymous

sorry can u plz elaborate ? where these 2 equations come from?

56. anonymous

Which one: (x-1)^1/3 - (x-3)^1/3 =1...........i (x-1)^2/3 +{(x-1)(x-3)}^1/3 + (x-3)^2/3=2........ii

57. anonymous

this one

58. anonymous

no second

59. anonymous

Second one just came from my mind

60. anonymous

ahh ok

61. anonymous

While I gave a thought when u said that x=2 only

62. anonymous

nope there is no answer for that system

63. anonymous

that gives$\sqrt{y}=-1$

64. anonymous

Yep

65. anonymous

So, Actually my approach was not really good............ As it may give somrtime solution when there should be no solution

66. anonymous

yeah maybe :)

67. anonymous

ok i really enjoyed guys....thank u

68. anonymous

Welcome........ And thanx for such a nice question

69. nipunmalhotra93

@sauravshakya haha yea we both had the same answer :D @mukushla I enjoyed it too man... thanks !!