Question
Decide whether the function $y=x^{3}-3 x^{2}+7$ is increasing or decreasing at $x = 1$ and $x = 3.$

Answer

$ y=x^{3}-3 x^{2}+7$
$ \therefore \frac{d y}{d x}=3 x^{2}-6 x$
$ \text { At } \boldsymbol{x}=\mathbf{1}$
$ \frac{d y}{d x}=3(1)^{2}-6(1)$
$ =3-6$
$ =-3<0 $
$\therefore \quad$ Function is decreasing at $x=1$.
$ \text { At } \boldsymbol{x}= \mathbf{3}$
$\frac{d y}{d x}= 3(3)^{2}-6(3)$
$= 27-18$
$= 9>0 $
$\therefore \quad$ Function is increasing at $x=3$.

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