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]}