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

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

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f(x)=2 and 1
how?
@nipunmalhotra93 @sauravshakya post complete solution ... i'll check it later

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Take it to the form a^3+b^3
|dw:1347716779899:dw|
Wll that be enough
So, f(x) can only be either 1 or 2
So the hard part is to find x
Was that correct @mukushla
x=2
thats correct...complete it :) can u find the exact range of f(x) ?
|dw:1347720322437:dw|
well, here's my solution. |dw:1347734655828:dw||dw:1347734803553:dw|
|dw:1347736319687:dw|
|dw:1347736404280:dw|
so three answers in total.... :)
Fantastic work @nipunmalhotra93 :)
i think the answer is only x=2
|dw:1347769714174:dw|
I think that also works
that gives f(x)=1 ?
YEP
And is solved the equation
emm... http://www.wolframalpha.com/input/?i=%5Csqrt%5B3%5D%7Bx-1%7D%2B%5Csqrt%5B3%5D%7B3-x%7D%3D1
WHAT! so my solution was wrong
because the range of that function is\[[\sqrt[3]{2},2]\]
Cant it be zero as x---->infinity
Oh I got it there is a problem with my solution
So if (x)^1/2 + y^1/2 =2........i and x^1/2 - y^1/2 =4.......ii
It has no solution........ though we get the value of x and y..... if we solve those equation
Same thing happened to my equations
right? @mukushla
where that equations come from?
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
Solved those two equations
Let x-1=A and x-3=B
Square first equation
|dw:1347770764883:dw|
@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
yeah thats right...have u checked ur solutions?
yup! when I plug in x=.98165, calculate cube root of 1-x and then put a minus sign with it. That works fine.
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.
@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
@mukushla can u PLZ CHECK it
I think the range of f(x) is|dw:1347775870157:dw|
thats right :)
good job saura :)
But I still have a question for u
guys @sauravshakya @nipunmalhotra93 u cooked the problem :D
yeah what ?
(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
There I solved for x and got the value
Similarly if I solved this (x)^1/2 + y^1/2 =2........i and x^1/2 - y^1/2 =4.......ii
Will I get the values of x and y
sorry can u plz elaborate ? where these 2 equations come from?
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
this one
no second
Second one just came from my mind
ahh ok
While I gave a thought when u said that x=2 only
nope there is no answer for that system
that gives\[\sqrt{y}=-1\]
Yep
So, Actually my approach was not really good............ As it may give somrtime solution when there should be no solution
yeah maybe :)
ok i really enjoyed guys....thank u
Welcome........ And thanx for such a nice question
@sauravshakya haha yea we both had the same answer :D @mukushla I enjoyed it too man... thanks !!

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