Cambridge International AS and A Level Mathematics Pure Mathematics 1

(Michael S) #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
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