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. $\left[\frac{2}{3}.2\right]$
• D. $(\frac{2}{3},2]$

Answer 1

The correct option is A $(\frac{2}{3},2)$

Let, $y=-x(x-2{)}^2=f(x)$

For a increasing function we know $f^{{}^{\prime }}\left(x\right)>0$

$\therefore f\left(x\right)=-x(x-2{)}^2$

differentiating we get,

$f^{{}^{\prime }}\left(x\right)=-2x(x-2)-(x-2{)}^2$

$f^{{}^{\prime }}\left(x\right)=-(x-2)(2x+x-2)$

$f^{{}^{\prime }}\left(x\right)=-(x-2)(3x-2)$

Now, as $f^{{}^{\prime }}\left(x\right)>0$ we have

$-(x-2)(3x-2)>0$

$\Rightarrow (x-2)(3x-2)<0$

$\Rightarrow x<2\text{or}x<\frac{2}{3}$

Therefore the interval is $(\frac{2}{3},2)$.