If the greatest and least values of the function $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-6{\mathrm{x}}^2+9\mathrm{x}+1$ on [0,2]  are $\mathrm{\lambda }\text{ and }\mathrm{\mu }\text{, }$ then the value of ${\mathrm{\lambda }}^4+{\mathrm{\mu }}^8$ must be

Given, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-6{\mathrm{x}}^2+9\mathrm{x}+1$

$\begin{array}{r}{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-12\mathrm{x}+9\\ =3(\mathrm{x}-1)(\mathrm{x}-3)\end{array}$

For maxima or minima f '(x) = 0

x =1, 3 but  $3\notin [0,2]$

Only one critical point in [0, 2] is x = 1

Greatest value 

$\begin{array}{r}\mathrm{\lambda }=max\left\{\mathrm{f}\right(0),\mathrm{f}(1),\mathrm{f}(2\left)\right\}\\ =max\{1,5,3\}=5\end{array}$

and least value

 $$\begin{gathered} \mu =min\{f\left(0\right),f\left(1\right),f\left(2\right)\} \\ =min \left\{1,5,3\right\}=1 \\ {\lambda }^4+{\mu }^8=5^4+1^8=625+1=626 \end{gathered}$$