Cambridge International AS and A Level Mathematics Pure Mathematics 1

Differentiation

148

P1^

5

For x = −1: x = − 2 ⇒ d

d

y

x

= 3(−2)^2 − 3 = + 9

x = 0 ⇒ d

d

y

x

= 3(0)^2 − 3 = −3.

For x = 1: x = 0 ⇒ d

d

y

x

= − 3

x = 2 ⇒ d

d

y

x

= 3(2)^2 − 3 = +9.

Thus the stationary point at x = −1 is a maximum and the one at x = 1 is a

minimum.

Substituting the x values of the stationary points into the original equation,

y = x^3 − 3 x + 1, gives

when x = −1, y = (−1)^3 − 3(−1) + 1 = 3

when x = 1, y = (1)^3 − 3(1) + 1 = −1.

There is a maximum at (−1, 3) and a minimum at (1, −1). The sketch can now be

drawn (see figure 5.19).

0

+ –

Figure 5.17

Figure 5.18^0

y

–1 1

minimum

(1, –1)

maximum

(–1, 3)

x

–1

3

0

1

Figure 5.19