Question

The interval of x for which the function $y=-x(x-2{)}^2$ increases is

• A. $[\frac{2}{3},2]$
• B. $[\frac{2}{3},2)$
• C. $(\frac{2}{3},2)$
• D. $(\frac{2}{3},2]$

Answer 1

The correct option is A $[\frac{2}{3},2]$

$y=f\left(x\right)=-x(x-2{)}^2$
$f^{\prime }\left(x\right)=-\left[2x\right(x-2)+(x-2{)}^2]$
$\Rightarrow f^{\prime }\left(x\right)=-3x^2+8x-4$
For increasing,
$f^{\prime }\left(x\right)\ge 0$
$\Rightarrow 3x^2-8x+4\le 0$
$\Rightarrow (x-\frac{2}{3})(x-2)\le 0$
$\Rightarrow x\in [\frac{2}{3},2]$