Question:
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 $x ?$
(a) $-3,-\frac{1}{3}$
(b) $-3, \frac{1}{3}$
(c) $3,-\frac{1}{3}$
(d) $3, \frac{1}{3}$
Solution:
(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 $x=\frac{1}{3}$