Find Positive Outputs: Solving y = -1/9x² + 1 2/3x for f(x) > 0

To solve $f(x) > 0$ for the function $f(x) = -\frac{1}{9}x^2 + 1\frac{2}{3}x$, follow these steps:

• Step 1: The quadratic equation is $-\frac{1}{9}x^2 + \frac{5}{3}x = 0$.
• Step 2: Factor out the greatest common factor: $x(-\frac{1}{9}x + \frac{5}{3}) = 0$.
• Step 3: Set each factor equal to zero: $x = 0$ and $-\frac{1}{9}x + \frac{5}{3} = 0$.
• Step 4: Solve for the second root: $x = \frac{5}{3} \times 9 = 15$.
• Step 5: The roots are $x = 0$ and $x = 15$. These divide the number line into intervals: $x < 0$, $0 < x < 15$, and $x > 15$.
Step 6: Test each interval to determine where $f(x)$ is positive:
• For $x < 0$, choose $x = -1$: $f(-1) = -\frac{1}{9}(-1)^2 + \frac{5}{3}(-1) = \frac{14}{9}$, so $f(x) > 0$.
• For $0 < x < 15$, choose $x = 1$: $f(1) = -\frac{1}{9}(1)^2 + \frac{5}{3}(1) = \frac{14}{9}$, so $f(x) > 0$.
• For $x > 15$, choose $x = 16$: $f(16) = -\frac{1}{9}(16)^2 + \frac{5}{3}(16) = -\frac{16}{9}$, so $f(x) \leq 0$.

The inequality $f(x) > 0$ holds for $x < 0$ or $x > 15$.

Therefore, the values of $x$ satisfying $f(x) > 0$ are $x > 15$ or $x < 0$.