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Find the maximum and minimum values of a y = 2x³+3x²-36x+1I need the answer quickly.. |
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Answer» Answer: y = 2x3 – 3x2 – 36x + 10 dydxdydx = 6X2 – 6x – 36 = 6(x2 – x – 6) dydxdydx = 0 gives 6(x2 – x – 6) = 0 6(x – 3) (x + 2) = 0 x = 3 (or) x = -2 d2ydx2d2ydx2 = 6(2x – 1) Case (i): when x = 3, (d2ydx2)x=3(d2ydx2)x=3 = 6(2 x 3 – 1) = 6 x 5 = 30, positive Since d2ydx2d2ydx2 is positive y is minimum when x = 3. The local minimum value is obtained by substituting x = 3 in y. Local minimum value = 2(33) – 3(32) – 36(3) + 10 = 2(27) – (27) – 108 + 10 = 27 – 98 = -71 Case (II): when x = -2, (d2ydx2)x=−2(d2ydx2)x=−2 = 6(-2 x 2 – 1) = 6 x -5 = -30, negative Since d2ydx2d2ydx2 is negative, y is maximum when x = -2. Local maximum value = 2(-2)3 – 3(-2)2 – 36(-2) + 10 = 2(-8) – 3(4) + 72 + 10 = -16 – 12 + 82 = -28 + 82 = 54Read more on Sarthaks.com - https://www.sarthaks.com/983672/find-the-local-minimum-and-local-maximum-of-y-2x-3-3x-2-36x-10 |
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