The equation of a curve is  d^2y / dx^2 = 2x-1. Given that the curve has a minimum point at (3, -10), find the coordinates of maximum point.?

d²y/dx² = 2x - 1
(d/dx)(dy/dx) = 2x - 1
d(dy/dx) = (2x - 1)dx
∫d(dy/dx) = ∫(2x - 1)dx
dy/dx = x² - x + C₁

The curve has a minimum point at x = 3:
dy/dx = 0 at x = 3
3² - 3 + C₁ = 0
C₁ = -6
Hence, dy/dx = x² - x - 6

dy/dx = x² - x - 6
dy = (x² - x - 6)dx
∫dy = ∫(x² - x - 6)dx
y = (x³/3) - (x²/2) - 6x + C₂

(3, -10) is a point at the curve:
-10 = (3³/3) - (3²/2) - 6(3) + C₂
C₂ = 7/2
Hence, y = (x³/3) - (x²/2) - 6x + (7/2)

When dy/dx = 0:
x² - x - 6 = 0
(x - 3)(x + 2) = 0

When x = -2:
y = [(-2)³/3] - [(-2)²/2] - 6(-2) + (7/2) = 65/6
dy/dx = 0
d²y/dx² = 2(-2) - 1 = -5 < 0

Hence, maximum point at (-2, 65/6).

y'' = 2x - 1
y' = x^2 - x + C

y' = 0 when x = 3

0 = 3^2 - 3 + C
0 = 9 - 3 + C
0 = 6 + C
-6 = C

y' = x^2 - x - 6
y = (1/3) * x^3 - (1/2) * x^2 - 6x + C

y = -10 when x = 3

-10 = (1/3) * 3^3 - (1/2) * 3^2 - 6 * 3 + C
-10 = 9 - 9/2 - 18 + C
18 - 10 - 9 + 9/2 = C
-1 + 4.5 = C
3.5 = C

y = (1/3) * x^3 - (1/2) * x^2 - 6x + (7/2)


y' = 0
0 = x^2 - x - 6
x = (1 +/- sqrt(1 + 24)) / 2
x = (1 +/- 5) / 2
x = -4/2 , 6/2
x = -2 , 3

y =>
(1/3) * (-2)^3 - (1/2) * (-2)^2 - 6 * (-2) + 7/2 =>
(1/3) * (-8) - (1/2) * (4) + 12 + 7/2 =>
-8/3 - 2 + 12 + 7/2 =>
10 + 21/6 - 16/6 =>
10 + 5/6 =>
60/6 + 5/6 =>
65/6

(-2 , 65/6) is the coordinate for the local maximum