Find the absolute maximum and absolute minimum of function $\displaystyle \mathrm{f}(x)=2 x^{3}-15 x^{2}+36 x+1$ on $\displaystyle [1,5]$.

\[\begin{aligned} & \mathrm{f}(x)=2 x^{3}-15 x^{2}+36 x+1 \\ & \Rightarrow \mathrm{f}^{\prime}(x)=6\left(x^{2}-5 x+6\right)=6(x-2)(x-3) \\ & \mathrm{f}^{\prime}(x)=0 \Rightarrow x=2,3 \in[1,5] \\ & \text { Now } \mathrm{f}(1)=24, \mathrm{f}(2)=29, \mathrm{f}(3)=28, \mathrm{f}(5)=56 \end{aligned} \] Hence, the absolute maximum value is $\displaystyle 56$ and the absolute minimum value is $\displaystyle 24$ .