(a) Find the maximum and minimum points of the... | WAEC SSCE - Further Mathematics - 2013

Question 1 Report

(a) Find the maximum and minimum points of the curve \(y = 2x^{3} - 3x^{2} - 12x + 4\).

(b) Sketch the curve in (a) above.

Answer

(a) Maximum and minimum points of \( y = 2x^{3} - 3x^{2} - 12x + 4 \).

At stationary points \( \dfrac{dy}{dx} = 0 \):

\[ \frac{dy}{dx} = 6x^{2} - 6x - 12 = 6\left(x^{2} - x - 2\right) = 6(x-2)(x+1). \]

Setting \( \dfrac{dy}{dx} = 0 \):

\[ 6(x-2)(x+1) = 0 \implies x = 2 \ \text{ or } \ x = -1. \]

The nature of each point is tested with the second derivative:

\[ \frac{d^{2}y}{dx^{2}} = 12x - 6. \]

At \( x = -1 \):

\[ \frac{d^{2}y}{dx^{2}} = 12(-1) - 6 = -18 < 0 \quad \Rightarrow \ \text{maximum}. \]\[ y = 2(-1)^{3} - 3(-1)^{2} - 12(-1) + 4 = -2 - 3 + 12 + 4 = 11. \]

Maximum point \( (-1,\ 11) \).

At \( x = 2 \):

\[ \frac{d^{2}y}{dx^{2}} = 12(2) - 6 = 18 > 0 \quad \Rightarrow \ \text{minimum}. \]\[ y = 2(2)^{3} - 3(2)^{2} - 12(2) + 4 = 16 - 12 - 24 + 4 = -16. \]

Minimum point \( (2,\ -16) \).

(b) Sketch of the curve.

To locate the curve, note the following guide points:

Feature	Point
Maximum turning point	\((-1,\ 11)\)
Minimum turning point	\((2,\ -16)\)
\(y\)-intercept \((x=0)\)	\((0,\ 4)\)
An \(x\)-intercept \((y=0)\)	\((-2,\ 0)\)

Since the leading coefficient is positive, the curve rises from the bottom-left, climbs to the maximum \((-1,\ 11)\), falls through the \(y\)-intercept \((0,\ 4)\) to the minimum \((2,\ -16)\), then rises again to the top-right, as shown below.

graph — Maximum at (-1, 11), minimum at (2, -16); y-intercept (0, 4).

Read lesson note on Polynomial Functions (WAEC)

Answer Details