(ii)

Solution:

(i)

Given function is $f(x)=3 x^{4}-8 x^{3}+12 x^{2}-48 x+25$ on $[0,3]$
On differentiating we get
$
\begin{array}{l}
f^{\prime}(x)=12 x^{3}-24 x^{2}+24 x-48 \\
f^{\prime}(x)=12\left(x^{3}-2 x^{2}+2 x-4\right) \\
f^{\prime}(x)=12(x-2)\left(x^{2}+2\right)
\end{array}
$
Now, for local minima and local maxima we have $f^{\prime}(x)=0$
$x=2$ or $x^{2}+2=0$ for which there are no real roots.
Therefore, we consider only $x=2 \in[0,3]$.
Then, we evaluate of $\mathrm{f}$ at critical point $\mathrm{x}=2$ and at the interval $[0,3]$
$
\begin{array}{l}
f(2)=3(2)^{4}-8(2)^{3}+12(2)^{2}-48(2)+25 \\
f(2)=48-64+48-96+25=-39 \\
f(0)=3(0)^{4}-8(0)^{3}+12(0)^{2}-48(0)+25=25 \\
f(3)=3(3)^{4}-8(3)^{3}+12(3)^{2}-48(3)+25=16
\end{array}
$
Hence, we can conclude that the absolute maximum value off on $[0,3]$ is 25 occurring at $x=0$ and the minimum value of $f$ on $[0,3]$ is $-39$ occurring at $x=2$

(ii)

And,
$
f(9)=(9-2) \sqrt{(9-1)}=7 \sqrt{8}=14 \sqrt{2}
$
Hence, we can conclude that the absolute maximum value of $\mathrm{f}$ is $14 \sqrt{2}$ occurring at $x=9$ and the minimum value of $f$ is $-2 \sqrt{3} / 9$ occurring at $x=4 / 3$.