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aditya jain

2x = y^1/5+ y ^(-1/5)

^{when will this function exist}

Aatish Pandit

its simply.......for all real values of y except ZERO...........

PLS RATE IF CORRECT

animesh prakash

Hey,
see it is simple.
it exists for y=R-{0} and for x=R-(-2,2).
SIMPLY,by A.M.>=G.M.
^{|(y^1/5 +y^-1/5)| /2 >= (y^1/5 * y^-1/5)}|(y^1/5 +y^-1/5)|/2 >= 1|(y^1/5 +y^-1/5)|>=2
i.e.|x|>=2^{i.e. x=R-(-2,2)
i have put modulus sign on}(y^1/5+y^-1/5)^{because A.M--G.M. theorem can be applied on positive nos. only.
also ,obyviously y can't be zero .

rate me if i was helpful}

HIMANSHU JAIN

look,

sum of ano. & it's reciprocal is always

greater than or equal to

2.

let y^1/5 = t

=> y^-1/5 = 1/t

2x = t+ 1/t

acc. to above statement

2x

2

x

1.

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