This graph is a straight line so you just need to use the extreme points.
x –2 4 g (x) –10 2
(c) f (x) = g (x) x 2 + 2x − 8 = 2x − 6 x 2 = 2 x 2 = ± √ ∴ √
__ 2 ≈ 1·4
__ 2
(b) Cubic curves: y = ax 3 + bx 2 + cx + d 1. Local maxima and local minima: y = ax 3 + bx 2 + cx + d dy
___ dx = 3 ax 2 + 2bx + c = 0
Solve this quadratic equation to get the stationary points. Test the stationary points in
(turning points).
___ dx 2 = 6ax + 2b to get the local maximum and local minimum
d 2 y 2. Crossing the axes:
(i) y-axis: x = 0: y = (0, d ) always (ii) x-axis: Solve y = 0 if you can. Shapes:
y x a > 0 EXAMPLE 16
(a) Find the co-ordinates of the local maximum and the local minimum of y = f (x) = 4 − 9x + 6x 2 − x 3 , x ∈ R. (b) Plot the curve in the domain −1 ≤ x ≤ 5, x ∈ R. (c) Use the graph to fi nd the range of values of x for which: (i) f (x) ≥ 0
(ii) f ʹ(x) < 0
Solution y = 4 − 9x + 6x 2 − x 3
(a)
___ dx = −9 + 12x − 3x 2
dy
−9 + 12x − 3x 2 = 0 x 2 − 4x + 3 = 0
(x − 1)(x − 3) = 0 ⇒ x = 1, 3
___ dx 2 = 12 − 6x
d 2 y
___ dx 2 = 12 – 6(1) = 6 > 0 There is a local minimum at x = 1. x = 1: y = 4 – 9(1) + 6(1)2 – (1)3 = 0 (1, 0) is a local minimum.
x = 1: d 2 y x = 3:
___ dx 2 = 12 – 6(3) = – 6 < 0
d 2 y
There is a local maximum at x = 3. x = 3: y = 4 – 9(3) + 6(3)2 – (3)3 = 4 (3, 4) is a local maximum.
