For what values of x is the rate of increase

(d) $3, \frac{1}{3}$

Let $y=x^{3}-5 x^{2}+5 x+8$

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

According to the question,

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

$\Rightarrow 3 x^{2}-10 x+5=2$

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

$\Rightarrow 3 x^{2}-9 x-x+3=0$

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

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

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