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

(A) Let $f(x) = 3x^{4}-8x^{3}+12x^{2}-48x+25$.
First,find the derivative $f'(x) = 12x^{3}-24x^{2}+24x-48$.
Set $f'(x) = 0$ to find critical points: $12(x^{3}-2x^{2}+2x-4) = 0$.
Factoring the cubic: $12[x^{2}(x-2)+2(x-2)] = 0$,which gives $12(x-2)(x^{2}+2) = 0$.
The only real critical point in the interval $[0,3]$ is $x=2$.
Now,evaluate $f(x)$ at the critical point and the endpoints of the interval:
At $x=0$: $f(0) = 3(0)^{4}-8(0)^{3}+12(0)^{2}-48(0)+25 = 25$.
At $x=2$: $f(2) = 3(2)^{4}-8(2)^{3}+12(2)^{2}-48(2)+25 = 3(16)-8(8)+12(4)-96+25 = 48-64+48-96+25 = -39$.
At $x=3$: $f(3) = 3(3)^{4}-8(3)^{3}+12(3)^{2}-48(3)+25 = 3(81)-8(27)+12(9)-144+25 = 243-216+108-144+25 = 16$.
Comparing these values: $25, -39, 16$.
The maximum value is $25$ at $x=0$ and the minimum value is $-39$ at $x=2$.