How do you find the domain and range of #root3( (x^2-x-12))#?

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Answer 1 · 2017-08-27T08:29:58

Domain: #(-oo, oo)#

Range: #[-root(3)(98)/2, oo)#

Given:

#f(x) = root(3)(x^2-x-12)#

We can convert the quadratic radicand into vertex form by completing the square:

#x^2-x-12 = x^2-2(1/2)x+(1/2)^2-(1/2)^2-12#

#color(white)(x^2-x-12) = (x-1/2)^2-(1/4+48/4)#

#color(white)(x^2-x-12) = (x-1/2)^2-49/4#

So the minimum value #-49/4# of the radicand occurs when #x=1/2#

Then:

#f(1/2) = root(3)(-49/4) = -root(3)(98/8) = -root(3)(98)/2#

Hence the range of #f(x)# is #[-root(3)(98)/2, oo)#.

Its domain is the whole of #RR = (-oo, oo)#, since it is well defined for any value of #x#.