Let #f: A rarr B# be an onto function such that #f(x) = sqrt(x-2-2sqrt(x-3)) - sqrt(x-2+2sqrt(x-3))#, then set 'B' is?

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Answer 1 · 2018-05-03T16:49:47

#B = {f in RR: -2 <= f <= 0}#

The set #A# is the domain and it is limited by the domain of #sqrt(x-3)#, therefore, set A is:

#A = {x in RR : x>=3}#

We can find one of the boundaries of set #B# by evaluating #f(3)#:

#f(3) = sqrt(3-2-2sqrt(3-3)) - sqrt(3-2+2sqrt(3-3))#

#f(3) = sqrt(1-2sqrt(0)) - sqrt(1+2sqrt(0))#

#f(3) = 0#

We find the other boundary of set #B# by finding the limit:

#lim_(x to oo) sqrt(x-2-2sqrt(x-3)) - sqrt(x-2+2sqrt(x-3))#

let #u = sqrt(x-3)#, then #u^2 = x-3# or #x-2 = u^2+1#

#lim_(u to oo) sqrt(u^2-2u+1) - sqrt(u^2+2u+1)#

The arguments are perfect squares:

#lim_(u to oo) sqrt((u-1)^2) - sqrt((u+1)^2)#

#lim_(u to oo) u-1 - (u+1) = -2#

The set B is:

#B = {f in RR: -2 <= f <= 0}#