How do you find the domain and range of #f(x)= sqrt(x^2-x-6)#?

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Answer 1 · 2016-06-17T05:47:26

Domain is #(-oo,-2[ uu]3,+oo)#
Range is #[0,+oo)#

Since under square root the polynomial must be positive, the domain is obtained by solving:

#x^2-x-6>=0#

you can obtain the zeroes of the polynomial by solving the associated equation:

#x^2-x-6=0#

#x=(1+-sqrt(1-4*(-6)))/2#

#x=(1+sqrt(25))/2#

#x=(1+-5)/2#

#x=-2 and x=3#

so the disequation is solved in the external intervals:

#x<-2 and x>3#
the the domain is:

#(-oo,-2[ uu]3,+oo)#

Since f(x) is positive, due to the result of square root, the range includes all positive real numbers: #x>=0#