question_answer 1)

Find both the maximum and minimum values of f(x) = 3x4 – 8x3+ 12x2 – 48x + 25 on the interval [0, 3}.

Answer:

f(x) = 3x4 ? 8x3 + 12x2 ? 48x + 25             = 12 [x3 ? 2x2 + 2x ? 4]       = 12 [x2 (x ? 2) + 2(x ? 2)]       = 12 (x ? 2) (x2 + 2)       f?(x) = 0

x	0	2	3
f(x)	25	-39	16

       Maximum value of f(x) = 25 at x = 0 and minimum value of (x) = ?39 at x = 2.