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

1 Answer
Apr 30, 2017

Answer:

Domain of #x# in interval notation is #[0,1)uu(1,oo)# and range is #RR#.

Explanation:

Here in #f(x)=sqrtx/(x^2+x-2)#

As we have #sqrtx# in numerator, #x# cannot take negative values, hence we should have #x>=0#

Further denominator can be factorized to #(x+2)(x-1)#, hence #x# cannot take values #-2# and #1#,

and hence domain of #x# in interval notation is #[0,1)uu(1,oo)#

However, as #f(x)# can take any value from #-oo# (when #x->1# from left) to #oo# (when #x->1# from right) as degree of denominator is higher than that of denominator. Further, #f(x)=0# when #x=0#..

hence range is #f(x) in RR#

graph{sqrtx/(x^2+x-2) [-7.58, 12.42, -4.96, 5.04]}