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

- anonymous

- schrodinger

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- anonymous

f(x)=2 and 1

- anonymous

how?

- anonymous

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

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## More answers

- anonymous

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

- anonymous

|dw:1347716779899:dw|

- anonymous

Wll that be enough

- anonymous

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

- anonymous

So the hard part is to find x

- anonymous

Was that correct @mukushla

- anonymous

x=2

- anonymous

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

- anonymous

|dw:1347720322437:dw|

- nipunmalhotra93

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

- nipunmalhotra93

|dw:1347736319687:dw|

- nipunmalhotra93

|dw:1347736404280:dw|

- nipunmalhotra93

so three answers in total.... :)

- anonymous

Fantastic work @nipunmalhotra93 :)

- anonymous

i think the answer is only x=2

- anonymous

|dw:1347769714174:dw|

- anonymous

I think that also works

- anonymous

that gives f(x)=1 ?

- anonymous

YEP

- anonymous

And is solved the equation

- anonymous

emm...
http://www.wolframalpha.com/input/?i=%5Csqrt%5B3%5D%7Bx-1%7D%2B%5Csqrt%5B3%5D%7B3-x%7D%3D1

- anonymous

WHAT! so my solution was wrong

- anonymous

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

- anonymous

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

- anonymous

Oh I got it there is a problem with my solution

- anonymous

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

- anonymous

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

- anonymous

Same thing happened to my equations

- anonymous

right? @mukushla

- anonymous

where that equations come from?

- 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

- anonymous

Solved those two equations

- anonymous

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

- anonymous

Square first equation

- anonymous

|dw:1347770764883:dw|

- 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

- anonymous

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

- 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.

- 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.

- 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

- anonymous

@mukushla can u PLZ CHECK it

- anonymous

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

- anonymous

thats right :)

- anonymous

good job saura :)

- anonymous

But I still have a question for u

- anonymous

guys @sauravshakya @nipunmalhotra93 u cooked the problem :D

- anonymous

yeah what ?

- 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

- anonymous

There I solved for x and got the value

- anonymous

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

- anonymous

Will I get the values of x and y

- anonymous

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

- 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

- anonymous

this one

- anonymous

no second

- anonymous

Second one just came from my mind

- anonymous

ahh ok

- anonymous

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

- anonymous

nope there is no answer for that system

- anonymous

that gives\[\sqrt{y}=-1\]

- anonymous

Yep

- anonymous

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

- anonymous

yeah maybe :)

- anonymous

ok i really enjoyed guys....thank u

- anonymous

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

- nipunmalhotra93

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

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