9.7 Second-Order Differentiation, Turning Points, Maximum and Minimum Points (Examples)

Example 1 (Maximum Value of Quadratic Function)

Given that y = 3x (4 – x), calculate

(a) the value of x when y is a maximum,

(b) the maximum value of y.

Solution:

(a)

\begin{array}{l} y = 3 x (4 - x) \\ y = 12 x - 3 x^{2} \\ \frac{d y}{d x} = 12 - 6 x \\ When y is maximum, \frac{d y}{d x} = 0 \\ 0 = 12 - 6 x \\ x = 2 \end{array}

(b)

\begin{array}{l} y = 12 x - 3 x^{2} \\ When x = 2, \\ y = 12 (2) - 3 {(2)}^{2} \\ y = 12 \end{array}

Example 2 (Determine the Turning Points and Second Derivative Test)

Find the coordinates of the turning points on the curve y = 2x³+ 3x²– 12x + 7 and determine the nature of these turning points.

Solution:

\begin{array}{l} y = 2 x^{3} + 3 x^{2} - 12 x + 7 \\ \frac{d y}{d x} = 6 x^{2} + 6 x - 12 \\ At turning point, \frac{d y}{d x} = 0 \end{array}

6x² + 6x – 12 = 0

x² + x – 2 = 0

(x – 1) (x + 2) = 0

x = 1 or x = –2

When x = 1

y = 2(1)³ + 3(1)² – 12(11) + 7

y = 0

(1, 0) is a turning point.

When x = –2

y = 2(–2)³ + 3(–2)² – 12(–2) + 7

y = 27
(–2, 27) is a turning point.

\begin{array}{l} \frac{d^{2} y}{d x^{2}} = 12 x + 6 \\ When x = 1, \\ \frac{d^{2} y}{d x^{2}} = 12 (1) + 6 = 18 > 0 (positive) \end{array}

Hence, the turning point (1, 0) is a minimum point.

\begin{array}{l} When x = - 2, \\ \frac{d^{2} y}{d x^{2}} = 12 (- 2) + 6 = - 18 < 0 (negative) \end{array}

Hence, the turning point (–2, 27) is a maximum point.