Find the value of x at the minimum point of the curve y = x3…

Find the value of x at the minimum point of the curve y = x³ + x² – x + 1

Correct Answer: Option D

Explanation

y = x³ + x² – x + 1
(frac{dy}{dx}) = (frac{d(x^3)}{dx}) + (frac{d(x^2)}{dx}) – (frac{d(x)}{dx}) + (frac{d(1)}{dx})
(frac{dy}{dx}) = 3x² + 2x – 1 = 0
(frac{dy}{dx}) = 3x² + 2x – 1
At the maximum point (frac{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 = (frac{1}{3}) or -1
For the maximum point
(frac{d^2y}{dx^2})

(frac{d^2y}{dx^2}) 6x + 2
when x = (frac{1}{3})
(frac{dx^2}{dx^2}) = 6((frac{1}{3})) + 2
= 2 + 2 = 4
(frac{d^2y}{dx^2}) > o which is the minimum point
when x = -1
(frac{d^2y}{dx^2}) = 6(-1) + 2
= -6 + 2 = -4
-4

therefore, (frac{d^2y}{dx^2})

the maximum point is -1