How do you find the domain and range of f(x) = sqrt x / (x^2 + x - 2)?

1 Answer
Dec 26, 2017

Domation: 0<=x<1 and x>1
Range: y in RR

Explanation:

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