Let $f(x)=\frac{x}{8*x^2+9}.$ Then

$f$ has a local minimum at $x=$

$f$ has a local maximum at $x=$

$2$

Let $f(x)=5*x^9-6*x^8$

Tick all of the following statements that are correct.

$f$ has a stationary point at $x=0.$

$f$ is increasing on $(0,1).$

If $x>\frac{6}{5}$ then $f(x)>f(\frac{6}{5}).$

$f$ is increasing on $(-,0).$

$f^{\text{'}}(x)>0$ for all $x>0$

$f$ is decreasing on $(\frac{16}{15},).$

$f$ is decreasing on $[0,).$

$f$ has a turning point at $x=\frac{16}{15},$