Find both the maximum value and the minimum value of $3{x^4} - 8{x^3} + 12{x^2} - 48x + 25$ on the interval $\left[ {0,3} \right]$.

Let $f\left( x \right) = 3{x^4} - 8{x^3} + 12{x^2} - 48x + 25,\,x \in \left[ {0,3} \right]$

$\Rightarrow f'\left( x \right) = 12{x^3} - 24{x^2} + 24x - 48 = 12\left\{ {{x^3} - 2{x^2} + 2x - 4} \right\}$
$= 12\left\{ {{x^2}\left( {x - 2} \right) + 2\left( {x - 2} \right)} \right\} = 12\left( {x - 2} \right)\left( {{x^2} + 2} \right)$

For critical points, $f'\left( x \right) = 0$
$\Rightarrow 12\left( {x - 2} \right)\left( {{x^2} + 2} \right) = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2 \in \left[ {0,3} \right]$

So, to find the maximum and minimum values, we have to evaluate $f\left( 0 \right),f\left( 3 \right)$ and $f\left( 2 \right)$.

Now, $f\left( 0 \right) = 25$
$f\left( 2 \right) = 3 \times {2^4} - 8 \times {2^3} + 12 \times {2^2} - 48 \times 2 + 25 = - 39$
$f\left( 3 \right) = 3 \times {3^4} - 8 \times {3^3} + 12 \times {3^2} - 48 \times 3 + 25 = 16$

Therefore the maximum value of $f\left( x \right)$ is $25$ at $x = 0$ and minimum value of $f\left( x \right)$ is $- 39$ at $x = 2$.