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Find the maximum and minimum values of 3x4 - 8x3+12x2 - 48x+1 on the interval [1, 4]. |
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Answer» 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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