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 $x$ ?
Answer & explanation
Correct answer: option 4
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
$\left(3 x^2-10 x+5\right) \frac{d x}{d t}=2 \frac{d x}{d t}$
$\Rightarrow 3 x^2-10 x+3=0$
$\Rightarrow (3 x-1)(x-3)=0$
$\Rightarrow x=3, \frac{1}{3}$