Find the maximum and minimum values of 
on the interval 
.
max. value is 257 at x = 4 and min. value is −63 at x = 2
F|(x)=12x3-24x2+24x-48=0
12(x3-2x2+2x-4)=0
Since for x=2, x3-2x2+2x-4=0, x-2 is a factor
On dividing x3-2x2+2x-4 by x-2, we get,
12(x-2)(x2+2)=0
X=2,4
Now, we shall evaluate the value of f at these points and the end points
F(1)=3(1)4-8(1)3+12(1)2-48(1)+1=-40
F(2)= 3(2)4-8(2)3+12(2)2-48(2)+1=-63
F(4)= 3(4)4-8(4)3+12(4)2-48(4)+1=257
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