Q. 16

# Find the ma

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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