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 $x$ .

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