For what values of $x$ is the rate of increase of $x^3-5 x^2+5 x+8$ is twice the rate of increase of $x$ ?

$3, \frac{1}{3}$

Let $y=x^3-5 x^2+5 x+8$. Then,

$\frac{d y}{d t}=\left(3 x^2-10 x+5\right) \frac{d x}{d t}$

When $\frac{d y}{d t}=2 \frac{d x}{d t}$, we have

$\left(3 x^2-10 x+5\right) \frac{d x}{d t}=2 \frac{d x}{d t}$

$\Rightarrow 3 x^2-10 x+3=0$

$\Rightarrow (3 x-1)(x-3)=0$

$\Rightarrow x=3, \frac{1}{3}$