Find the interval in which the function f (x)=x^4-4 x^3+10 is strictly decreasing.

$$\begin{aligned} & \mathrm{f}^{\prime}(x)=$\displaystyle 4$ x^{$\displaystyle 3$}-$\displaystyle 12$ x^{$\displaystyle 2$}=$\displaystyle 4$ x^{$\displaystyle 2$}(x-$\displaystyle 3$) & \Rightarrow $\displaystyle 4$ x^{$\displaystyle 2$}(x-$\displaystyle 3$)<$\displaystyle 0$ \text { for } x<$\displaystyle 3$, x \neq $\displaystyle 0$ & \Rightarrow \mathrm{f}^{\prime}(x)<$\displaystyle 0$ \text { for } x<$\displaystyle 3$, x \neq $\displaystyle 0$ \end{aligned}

$$Thus, $\displaystyle \mathrm{f}(x)=x^{4}-4 x^{3}+10$ is strictly decreasing on $\displaystyle (-\infty, 0) \cup(0,3)$.

Thus, $\displaystyle \mathrm{f}(x)=x^{4}-4 x^{3}+10$ is strictly decreasing on $\displaystyle \left.(-\infty, 0] \cup[0,3]\right)$