Cambridge International AS and A Level Mathematics Pure Mathematics 1

Differentiation

156

P1^

5

The next example illustrates the use of the second derivative to identify the

nature of stationary points.

EXAMPLE 5.15 Given that y = 2 x^3 + 3 x^2 − 12 x

(i) find d

d

y

x

, and find the values of x for which d

d

y

x

= 0

(ii) find the value of d

d

2

2

y

x

at each stationary point and hence determine its nature

(iii) find the y values of each of the stationary points

(iv) sketch the curve given by y = 2 x^3 + 3 x^2 − 12 x.

SOLUTION

(i) d

d

y

x

= 6 x^2 + 6 x − 12

= 6(x^2 + x − 2)

= 6(x + 2)(x − 1)

d

d

y

x

= 0 when x = −2 or x = 1.

(ii) d

d

2

2

y

x

= 12 x + 6.

When x = −2, d

d

2

2

y

x

= 12 × (−2) + 6 = −18.

d

d

2

2

y

x

 0 ⇒ a maximum.

When x = 1, d

d

2

2

y

x

= 6(2 × 1 + 1) = 18.

d

d

2

2

y

x

 0 ⇒ a minimum.

(iii) When x = −2, y = 2(−2)^3 + 3(−2)^2 − 12(−2)

= 20

so (−2, 20) is a maximum point.

When x = 1, y = 2 + 3 − 12

= − 7

so (1, −7) is a minimum point.

0

P

+ –

d^2 y

dx^2

< 0

at P

Figure 5.26 Figure 5.27

0

Q

d^2 y

dx^2

> 0

at Q