Question #4fbab

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Answer 1 · 2017-03-16T20:04:47

The domain of $f(x)$ is $x < -4 and x > 3$

Because the argument for the $1/2$ power cannot be negative we must find the part of the domain where $x^2+x-12>=0$ and exclude it.

Let's find the points where the argument is 0:

$x^2+x-12=0$

Factor:

$(x +4)(x-3)=0$

This implies that:

$x+4=0 and x-3=0$

$x=-4 and x=3$

The function is negative between these values of x:

graph{x^2+x-12 [-8, 8, -14, 14]}

Therefore, the domain of $f(x)$ is $x < -4 and x > 3$