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