2003
Mathematics
67ca00ce0c643c71d77dee1f
Determine the maximum value of y = 3x2 - x3
A.
zero
B.
2
C.
4
D.
6
Correct Answer: 4
Explanation
y = 3x2 - x3 dy/dx = 6x - 3x2 as dy/dx = 0 6x - 3x2 = 0 3x (2 - x) = 0 this implies that 2 -x = 0 and 3x = 0 x = 2 (or) 0 But = dy/dx = 6x - 3x2 d2y/dx2 = 6 - 6x at x = 2 = 6 - 6(2) = -6 y = 3x2 - x3 = 3(2)2 - 23 = 12 - 8 = 4
y = 3x2 - x3 dy/dx = 6x - 3x2 as dy/dx = 0 6x - 3x2 = 0 3x (2 - x) = 0 this implies that 2 -x = 0 and 3x = 0 x = 2 (or) 0 But = dy/dx = 6x - 3x2 d2y/dx2 = 6 - 6x at x = 2 = 6 - 6(2) = -6 y = 3x2 - x3 = 3(2)2 - 23 = 12 - 8 = 4THIS WEEK's
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