What is the domain and range of $f(x) = sqrt x /( x^2 + x - 2)$ ?

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Answer 1 · 2018-01-06T13:26:36

The domain is $x in [0,1)uu(1,+oo)$ .
The range is $y in RR$

The function is

$f(x)=sqrt(x)/(x^2+x-2)$

Factorise the denominator

$(x^2+x-2)=(x+2)(x-1)$

From the $sqrt$ sign , $x>=0$

Therefore,

The domain is

$x in [0,1)uu(1,+oo)$

Let,

$y=sqrt(x)/((x+2)(x-1))$

When $x=0$ , $f(0)=0$

$lim_(x->1^-)sqrt(x)/((x+2)(x-1))=1/0^-=-oo$

$lim_(x->1^+)sqrt(x)/((x+2)(x-1))=1/0^+=+oo$

The range is $y in RR$

graph{sqrt(x)/(x^2+x-2) [-8.89, 8.89, -4.444, 4.445]}

What is the domain and range of f(x) = sqrt x /( x^2 + x - 2)f(x) = sqrt x /( x^2 + x - 2)?