Solve y = (1/3)x² + (2/3)x: Finding Values Where Function is Negative

To solve the problem, we follow these steps:

• Step 1: Identify the quadratic equation $\frac{1}{3}x^2 + \frac{7}{3}x = 0$.
• Step 2: Factor the equation $x(\frac{1}{3}x + \frac{7}{3}) = 0$.
• Step 3: Set each factor to zero to find the roots: $x = 0$ and $\frac{1}{3}x + \frac{7}{3} = 0$.
• Step 4: Solve for $x$ to find the second root: $x = -7$.
• Step 5: Analyze the intervals: Plug in test points from intervals defined by $x = -7$ and $x = 0$ to determine sign of $f(x)$.

Now, let's solve each step:

Step 1: The function is $\frac{1}{3}x^2 + \frac{7}{3}x = 0$.
Step 2: Factor the equation: as $x(\frac{1}{3}x + \frac{7}{3}) = 0$. Thus, $x = 0$ and $\frac{1}{3}x + \frac{7}{3} = 0$ are the potential roots.
Step 3: Solve for the second root $x$:

$\frac{1}{3}x + \frac{7}{3} = 0$

Multiply through by 3 to clear fractions:

$x + 7 = 0$

Solve for $x$:
$x = -7$.

Step 4: Identify intervals of interest: We have critical points at $x = -7$ and $x = 0$. Thus, we have intervals $(- \infty, -7)$, $(-7, 0)$, and $0, \infty)$.
Step 5: Test these intervals to determine where $\frac{1}{3}x^2 + \frac{7}{3}x < 0$.

Choose test values:
- For interval $(-8): (-8, 0) \rightarrow \frac{1}{3}(-8)^2 + \frac{7}{3}(-8) = \frac{64}{3} - \frac{56}{3} = \frac{8}{3} > 0$, - For interval $(-2): (-2, 0) \rightarrow \frac{1}{3}(-2)^2 + \frac{7}{3}(-2) = \frac{4}{3} - \frac{14}{3} = -\frac{10}{3} < 0$.

Conclusively, the solution is:

$-7 < x < 0$