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Earlier discussion
1 older comment kept for context.
Oluwaseun Adeyemi
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 maximum point is -1