The global maximum and minimum values of the continuous function #f# on the closed and bounded (compact) interval #[0,3]# will occur either at the critical points inside the interval or at the endpoints.
The derivative is #f'(x)=12x^2-52x+16=4(3x^2-13x+4)=4(3x-1)(x-4)#, which leads to critical points at #x=1/3# and #x=4#. Only #x=1/3# is in the interval #[0,3]#, so that's the only one we need to consider.
Now #f(0)=1#, #f(1/3)=4\cdot 1/27-26\cdot 1/9+16\cdot 1/3+1=97/27\approx 3.6#, and #f(3)=108-234+48+1=-77#.
Therefore, the global maximum value of #f# on #[0,3]# is #97/27# at #x=1/3# and the global minimum value of #f# on #[0,3]# is #-77# at #x=3#.
