What are the stationary points of the curve (1/3)x^3 - 2x^2 + 3x + 2 and what is the nature of each stationary point.

Firstly, a stationary point is a point on the curve where the gradient of the tangent line is equal to 0. Here, this occurs at points where the first derivative of f(x) = 1/3x³ - 2x² + 3x + 2 is equal to zero. The first derivative is given by f'(x) = (1/3)3x² - 22x + 31 + 0, which is obtained using the fact that the derivative of axⁿ is anx^n-1. Simplifying f'(x) gives f'(x) = x²- 4x +3. Setting this derivative to 0, we obtain 0 = x² - 4x + 3. Factorising: 0 = (x - 1)(x - 3). Thus the first derivative is equal to 0 at x=1 and x=3. Substituting these x-values into the initial curve gives f(1) = 1/3 * 1³ - 21² + 31 + 2 = 10/3 and f(3) = 1/3 * 3³ - 23² + 3*3 + 2 = 2. Thus the stationary points of the curve are (1, 10/3) and (3, 2). To determine their nature, we consider the second derivative of the curve. This is given by f''(x) = 2x - 4. If the second derivative of the curve, evaluated at the x-value of a stationary point, is positive, then the stationary point is a minimum. If it is negative, then the stationary point is a maximum. For (1, 10/3), f''(1) = -2 < 0 so (1, 10/3) is a maximum. For (3, 2), f''(3) = 2 > 0 so (3, 2) is a minimum. Stationary points are (1, 10/3) and (3, 2), which are a maximum and a minimum respectively.