Answer :

Let f (x) = 3x4 – 8x3 + 12x2 – 48x + 25, x [0,3]

f’(x) = 12x3 - 24x2 + 24x – 48


=12(x3 - 2x2 + 2x – 4)


=12[x2 (x – 2)+ 2(x – 2)]


=12(x -2)( x2+ 2)


Now, f’(x) = 0


x =2 or (x2+ 2) = 0 for which there are no real roots.


Therefore, we will only consider x = 2


Now, we evaluate the value of f at critical point x = 2 and at end points of the interval [0,3].


f(2) = 3(2)4 – 8(2)3 + 12(2)2 – 48(2) + 25


= 3(16) – 8(8) + 12(4) +25


=48 – 64 +48 – 96 + 25


= -39


f(0) = 3(0)4 – 8(0)3 + 12(0)2 – 48(0) + 25


= 0+ 0 + 0 +25


= 25


f(3) = 3(3)4 – 8(3)3 + 12(3)2 – 48(3) + 25


= 3(81) – 8(27) + 12(9) +25


=243 – 216 +108 – 144 + 25


= 16


Therefore, we have the absolute maximum value of f on [0,3] is 25 occurring at x =0.


And, the absolute minimum value of f on [0,3] is -39 occurring at x = 2.


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