Analyze Domain and Positivity of y = x² - 4/9: Complete Function Study

The function we are given is $y = x^2 - \frac{4}{9}$. This is a quadratic function.

To find where $f(x) > 0$, we first need to determine where the function equals zero and changes sign. This involves solving the equation:

$x^2 - \frac{4}{9} = 0$

Rearranging gives:

$x^2 = \frac{4}{9}$

Taking the square root of both sides, we find:

$x = \pm \frac{2}{3}$

These are the points where the function changes signs. The parabola represented by this quadratic function opens upwards (since the coefficient of $x^2$ is positive and equal to 1), indicating that it is positive outside the interval between these roots and negative inside:

• The intervals of interest are $x < -\frac{2}{3}$, $-\frac{2}{3} < x < \frac{2}{3}$, and $x > \frac{2}{3}$.
• In the interval $x < -\frac{2}{3}$, $x^2 - \frac{4}{9} > 0$ because outside the roots, the parabola is above the x-axis.
• In the interval $-\frac{2}{3} < x < \frac{2}{3}$, $x^2 - \frac{4}{9} < 0$ because it is between the roots where the parabola lies below the x-axis.
• In the interval $x > \frac{2}{3}$, $x^2 - \frac{4}{9} > 0$ for the same reason as $x < -\frac{2}{3}$.

Therefore, the function $f(x) > 0$ when $x < -\frac{2}{3}$ or $x > \frac{2}{3}$.

Considering the choices provided, the correct answer that satisfies $f(x) > 0$ is choice 3: $x > \frac{2}{3}$ or $x < -\frac{2}{3}$.