2011
Mathematics
JAMB
Find the value of x at the minimum point of the curve y = x3 + x2 - x + 1
A.
1/3
B.
1/3
C.
1
D.
-1
Correct Answer: 1/3
Explanation
y = x3 + x2 - x + 1 dy/dx = d(x3)/dx + d(x2)/dx - d(x)/dx + d(1)/dx dy/dx = 3x2 + 2x - 1 = 0 dy/dx = 3x2 + 2x - 1 At the maximum point dy/dx = 0 3x2 + 2x - 1 = 0 (3x2 + 3x) - (x - 1) = 0 3x(x + 1) -1(x + 1) = 0 (3x - 1)(x + 1) = 0 therefore x = 1/3 or -1 For the maximum point d2y/dx2 < 0 d2y/dx2 6x + 2 when x = 1/3 dx2/dx2 = 6(1/3) + 2 = 2 + 2 = 4 d2y/dx2, > o which is the minimum point when x = -1 d2y/dx2 = 6(-1) + 2 = -6 + 2 = -4 -4 < 0 therefore, d2y/dx2 < 0 the minimum point is 1/3
y = x3 + x2 - x + 1 dy/dx = d(x3)/dx + d(x2)/dx - d(x)/dx + d(1)/dx dy/dx = 3x2 + 2x - 1 = 0 dy/dx = 3x2 + 2x - 1 At the maximum point dy/dx = 0 3x2 + 2x - 1 = 0 (3x2 + 3x) - (x - 1) = 0 3x(x + 1) -1(x + 1) = 0 (3x - 1)(x + 1) = 0 therefore x = 1/3 or -1 For the maximum point d2y/dx2 < 0 d2y/dx2 6x + 2 when x = 1/3 dx2/dx2 = 6(1/3) + 2 = 2 + 2 = 4 d2y/dx2, > o which is the minimum point when x = -1 d2y/dx2 = 6(-1) + 2 = -6 + 2 = -4 -4 < 0 therefore, d2y/dx2 < 0 the minimum point is 1/3THIS WEEK's
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