Correct Answer: Option A

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

\(\frac{d^2y}{dx^2}\) < 0

\(\frac{d^2y}{dx^2}\) 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 < 0

therefore, \(\frac{d^2y}{dx^2}\) < 0

the minimum point is 1/3

There is an explanation video available below.

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