Domain Analysis: Solve (x+1/6)(-x-4/9) > 0 Step by Step

Find the positive and negative domains of the function below:

$y=\left(x+\frac{1}{6}\right)\left(-x-4\frac{1}{9}\right)$

Then determine for which values of $x$ the following is true:

$f\left(x\right) > 0$

To solve this problem, we will determine where the function, given as $y = \left(x + \frac{1}{6}\right)\left(-x - 4\frac{1}{9}\right)$, is positive.

Step 1: **Find the Roots**
Set the function equal to zero: $\left(x + \frac{1}{6}\right)\left(-x - 4\frac{1}{9}\right) = 0$. This yields:
- $x + \frac{1}{6} = 0$ which gives $x = -\frac{1}{6}$, and
- $-x - 4\frac{1}{9} = 0$ which gives $x = -4\frac{1}{9}$.
Thus, the roots are $x = -\frac{1}{6}$ and $x = -4\frac{1}{9}$.

Step 2: **Analyze Sign Intervals**
The parabola opens downwards because the product has a negative coefficient as the leading term.
We have intervals: $x < -4\frac{1}{9}$, $-4\frac{1}{9} < x < -\frac{1}{6}$, and $x > -\frac{1}{6}$.

Since the quadratic opens downwards, it is positive between the roots (-4\frac{1}{9}, -\frac{1}{6}), where $f(x) > 0$.

Therefore, the solution for the values of $x$ for which $f(x) > 0$ is:

$-4\frac{1}{9} < x < -\frac{1}{6}$