2.11.1 Differentiation Long Questions (Question 1)

Question 1:

The curve y = x³ – 6x² + 9x + 3 passes through the point P (2, 5) and has two turning points, A (3, 3) and B.

Find

(a) the gradient of the curve at P.

(b) the equation of the normal to the curve at P.

(c) the coordinates of B and determine whether B is a maximum or the minimum point.

Solution:

(a)

y = x³ – 6x² + 9x + 3

dy/dx = 3x²– 12x + 9

At point P (2, 5),

dy/dx = 3(2)² – 12(2) + 9 = –3

Gradient of the curve at point P = –3.

(b)

Gradient of normal at point P = 1/3

Equation of the normal at P (2, 5):

y – y₁= m (x – x₁)

y – 5 = 1/3 (x – 2)

3y – 15 = x – 2

3y = x + 13

(c)

At turning point, dy/dx = 0.

3x² – 12x + 9 = 0

x² – 4x + 3 = 0

(x – 1)( x – 3) = 0

x – 1 = 0 or x – 3 = 0

x = 1 x = 3 (Point A)

Thus at point B:

x = 1

y = (1)³– 6(1)² + 9(1) + 3 = 7

Thus, coordinates of B = (1, 7)

\begin{array}{l} when x = 1, \\ \frac{d^{2} y}{d x^{2}} = 6 x - 12 \\ \frac{d^{2} y}{d x^{2}} = 6 (1) - 12 \\ \frac{d^{2} y}{d x^{2}} = - 6 < 0 \\ Since \frac{d^{2} y}{d x^{2}} < 0, B is a maximum point . \end{array}