Cambridge International AS and A Level Mathematics Pure Mathematics 1

P1^

5

Maximum

(^) and
(^) minimum
(^) points
In this case you knew the general shape of the cubic curve and the positions of all
of the maximum and minimum points, so it was easy to select values of x for
which to test the sign of d
d
y
x

. The curve of a more complicated function may have
several maxima and minima close together, and even some points at which the
gradient is undefined. To decide in such cases whether a particular stationary
point is a maximum or a minimum, you must look at points which are just either
side of it.

EXAMPLE 5.11 Find all the stationary points on the curve of y = 2 t^4 − t^2 + 1 and sketch the curve.

SOLUTION

d

d

y

t

= 8 t^3 − 2 t

At a stationary point, d

d

y

t

= 0, so

8 t^3 − 2 t = 0

2 t(4t^2 − 1) = 0

2 t(2t − 1)(2t + 1) = 0

⇒ d

d

y

t

= 0 when t = −0.5, 0 or 0.5.

You may find it helpful to summarise your working in a table like the one below.

You can find the various signs, + or −, by taking a test point in each interval, for

example t = 0.25 in the interval 0  t  0.5.

t  −0.5 −0.5 −0.5  t  0 0 0  t  0.5 0.5 t  0.5

Sign of

d

dt

y − 0 + 0 − 0 +

Stationary point min max min

There is a maximum point when t = 0 and there are minimum points when

t = −0.5 and +0.5.

When t = 0: y = 2(0)^4 − (0)^2 + 1 = 1.

When t = −0.5: y = 2(−0.5)^4 − (−0.5)^2 + 1 = 0.875.

When t = 0.5: y = 2(0.5)^4 − (0.5)^2 + 1 = 0.875.

Therefore (0, 1) is a maximum point and (−0.5, 0.875) and (0.5, 0.875) are minima.