Solution

To solve this question is effectively the same as solving any inequality. In this case, since the directions reference "positive", we are solving \(f(x) \gt 0\). Note that 0 is *not* included.

Now, there is a "trick" to this problem which makes this a tiny bit easier: the denominator \(27(3x+4)^2\) is always positive! We still need to note that there is a domain restriction at \(x= -\frac{4}{3}\), but otherwise we are done with the denominator.

What remains now is to find the zeros from the numerator and then to identify the possible intervals for the solution: \[ \solve{ x^2+x-12 &=&0\\ (x-3)(x+4)&=&0 } \] So, \(x=-4\) and \(x=3\) are the zeros while \(x=-\frac{4}{3}\) is the domain restriction. Thus, the intervals of consideration are \((-\infty, -4)\),\(\left(-4,-\frac{4}{3}\right)\), \(\left(-\frac{4}{3},3\right)\), and \((3,\infty)\). After applying test values and checking for positive outcomes, you should determine that the intervals where the function is positive are \((-\infty,-4)\cup(3,\infty)\).

Finally, we can see this visually when we consider the graph of the function (though the exact answers may not necessarily be so easily determined from a graph on all questions).

f(x) = \frac{x^2+x-12}{27(3x+4)^2}\ f(x)\gt0\left\{0\leq y\right\}; {color:red}

Bonus Answer

This function is negative on \(\left(-4,-\frac{4}{3}\right)\cup\left(-\frac{4}{3},3\right)\). Notice that we *still* need to exclude the domain restriction!!