Discussion Forum

The local maximum of $y=x^{3}-3x^{2}+5$ is attained at 

Answer:

Explanation:

Given , $y=x^{3}-3x^{2}+5$  .........(i)

 On differentiating  both sides w.r.t 'x' , we get

 $\frac{dy}{dx}=3x^{2}-6x$  ........(ii)

 For local maxima or local minima , put $\frac{dy}{dx}=0$

$\Rightarrow$   $3x^{2}-6x=0$

$\Rightarrow$   $3x(x-2)=0$

$\Rightarrow$      x=0 or x=2

Now, differentiating Eq.(ii)  w.r.t .'x' we get

    $\frac{d^{2}y}{dx^{2}}=6x-6$

$\Rightarrow$        $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0}=-6<0$

$\therefore$   x=0  is a point of local maxima

and   $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=2}=6\times2-6=12-6=6>0$

 $\therefore$    x=2 is a point of local minima