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

Given, $y=x^3-3 x^2+5$ On differentiating both sides w.r.t. ' $x$ ', we get $$ \frac{d y}{d x}=3 x^2-6 x $$ For local maxima or local minima, put $\frac{d y}{d x}=0$ $$ \begin{aligned} & \Rightarrow & 3 x^2-6 x & =0 \\ & \Rightarrow & 3 x(x-2) & =0 \\ & \Rightarrow & x & =0 \text { or } x=2 \end{aligned} $$ Now, differentiating Eq. (ii) w.r.t. ' $x$ ' we get $$ \begin{aligned} & \frac{a^2 y}{d x^2}=6 x-6 \\ & \Rightarrow \quad\left(\frac{a^2 y}{d x^2}\right)_{x=0}=-6 < 0 \\ & \therefore x=0 \text { is a point of local maxima. } \\ & \text { and }\left(\frac{a^2 y}{d x^2}\right)_{x=2}=6 \times 2-6=12-6=6>0 \\ & \therefore x=2 \text { is a point of local minima. } \end{aligned} $$