Solve x²/5 + 1⅓x > 0: Finding Positive Function Values

Look at the following function:

$y=\frac{1}{5}x^2+1\frac{1}{3}x$

Determine for which values of $x$ the following is true:

$f(x) > 0$

Let's solve the problem step by step:

The function given is $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$. The first step is to convert the mixed number into an improper fraction:

$1\frac{1}{3} = \frac{4}{3}$, so the function becomes:

$y = \frac{1}{5}x^2 + \frac{4}{3}x$.

This can be written in standard form $ax^2 + bx + c = 0$ where $a = \frac{1}{5}$, $b = \frac{4}{3}$, and $c = 0$ (not visible, but necessary for proper representation).

Next, we use the quadratic formula to find the roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Plug in the values:

$x = \frac{-\frac{4}{3} \pm \sqrt{\left(\frac{4}{3}\right)^2 - 4 \times \frac{1}{5} \times 0}}{2 \times \frac{1}{5}}$.

Simplify where possible:

$x = \frac{-\frac{4}{3} \pm \sqrt{\frac{16}{9}}}{\frac{2}{5}}$, leading to:

$x = \frac{-\frac{4}{3} \pm \frac{4}{3}}{\frac{2}{5}}$.

This gives roots:

$x = 0$ and $x = -6\frac{2}{3}$.

The parabola opens upward (since $a = \frac{1}{5} > 0$). The quadratic function is positive between the roots and for values outside them:

This results in two intervals where $f(x) > 0$:

• $x < -6\frac{2}{3}$
• $x > 0$

Thus, the solution to the problem is:

$x > 0$ or $x < -6\frac{2}{3}$.