#f(x)=sqrtx/{x^2+x-2}=sqrtx/{(x+2)(x-1)}#
Domation:
#x>=0 and x!=-2,1 iff 0<=x<1 and x>1#
Range:
#f'(x)={1/{2sqrtx}(x^2+x-2)-sqrtx(2x+1)}/{[(x+2)(x-1)]^2}#
(note!! #f'(0)=NaN#)
let #f'(x)=0 => {x^2+x-2}/{2sqrtx}-sqrtx(2x+1)=0#
#iff {x^2+x-2}/{2sqrtx}=sqrtx(2x+1)#
#iff x^2+x-2=2x(2x+1)#
#iff x^2+x-2=4x^2+2x#
#iff 3x^2+x+2=0#
#=>#
there is no solution for this (I take #x in RR#)
#=>#
There aren't any max or min
#=>#
We will check what happens in each range (#0<=x<1 or x>1#)
#=>#
#f'(1/2)={1/{2sqrt{1/2}}((1/2)^2+1/2-2)-sqrt{1/2}(2(1/2)+1)}/{[(1/2+2)(1/2-1)]^2}=#
#={sqrt2/4(-5/4)-sqrt{1/2}(2)}/{(+)}=#
#=-{5sqrt2}/16-{16sqrt2}/16=#
#={-21sqrt2}/16=(-)<0#
#f'(4)={1/{2sqrt{4}}((4)^2+4-2)-sqrt{4}(2(4)+1)}/{[(4+2)(4-1)]^2}=#
#={1/4(18)-(2)(9)}/{(+)}=#
#2*9=18 , 18/4<18 => 1/4(18)-(2)(9)<0#
#=(-)<0#
Well, it was a long algebra but in the end we know now that #f(x)# goes down for all the range #0<=x<1 and x>1#
Now let's do simpler algebra:
#f(0)=0#
#lim_{x rarr 1^-}f=-oo#
#lim_{x rarr 1^+}f=+oo#
#lim_{x rarr oo}f=0^+#
SO:
#0<=y<-oo and 0>y iff y in RR#
