Let 
on [0, 3]
Now 
or 
Since is imaginary, therefore it is rejected.
is turning point.
At 
= 
At 
At 
Therefore, absolute minimum value is and absolute maximum value is 25.
Find both the maximum value and the minimum value of 3x ^4 – 8x^ 3 + 12x ^2 – 48x + 25 on the interval [0, 3].
Find both the maximum value and the minimum value of 3x ^4 – 8x^ 3 + 12x ^2 – 48x + 25 on the interval [0, 3].