$f(x)$ is a quadratic expression such that $f(x)$ is negative when $x \in\left(-\infty,-\frac{5}{3}\right)…
$f(x)$ is a quadratic expression such that $f(x)$ is negative when $x \in\left(-\infty,-\frac{5}{3}\right) \cup(3, \infty)$ and positive when $x \in\left(-\frac{5}{3}, 3\right) \cdot g(x)$ is another quadratic expression such that $g(x)$ is negative when $x \in\left(3, \frac{9}{2}\right)$ and positive when $x \in R-\left[3, \frac{9}{2}\right]$. Then, the sign of $f(x) g(x)$ in $[0,5]$ is
positive in $\left[0, \frac{9}{2}\right]$ and negative in $\left(\frac{9}{2}, 5\right)$
positive in $[0,3) \cup\left(3, \frac{9}{2}\right)$ and negative in $\left(\frac{9}{2}, 5\right]$
positive in $[0,3) \cup\left(3, \frac{9}{2}\right) \cup\left(\frac{9}{2}, 5\right]$
negative in $[0,3) \cup\left(3, \frac{9}{2}\right) \cup\left(\frac{9}{2}, 5\right]$
Solution
$f(x)=a\left(x+\frac{5}{3}\right)(3-x), a>0$ and $g(x)=b(x-3)\left(x-\frac{9}{2}\right), b>0$ Now, $f(x) \cdot g(x)=a b$
$
\left(x+\frac{5}{3}\right)(3-x)(x-3)\left(x-\frac{9}{2}\right)
$
According to wavy curve method,
So, $f(x) \cdot g(x)$ is positive in $[0,3) \cup\left(3, \frac{9}{2}\right)$ and negative in $\left(\frac{9}{2}, 5\right)$
