The Chemistry Maths Book, Second Edition

4.10 Stationary points 115

gradient decreases from positive values, through zero at A, to negative values. It

follows that the rate of change of the gradient is negative at A, and this is a sufficient

condition for the function to have maximum value at this point:

for a maximum: (4.21)

Similar considerations applied to the minimum at B show that

for a minimum: (4.22)

For the cubic shown in Figure 4.9,

y 1 = 1 x(x 1 − 1 3)

2

1 = 1 x

3

1 − 16 x

2

1 + 19 x

The point C, atx 1 = 12 in Figure 4.9, is an example of a point of inflection, at which the

gradientis a maximum or minimum, with The slope of the curve decreases

(becomes more negative) between A and C and increases between C and B, with

minimum value at C. This is an example of a simple point of inflection with

0 Exercises 78 – 82

When at a point then the nature of the point is determined by

the first nonzero higher derivative. Two examples are

(i) (4.23)

This is a point of inflection which is also a stationary point (but not a turning point),

and is the case discussed in Example 4.23.

(ii) (4.24)

dy

dx

dy

dx

dy

dx

dy

dx

=, 0000 =, =, ≠

2

2

3

3

4

4

dy

dx

dy

dx

dy

dx

=, 000 =, ≠

2

2

3

3

dy

dx

and

dy

dx

== 00

2

2

dy

dx

dy

dx

≠= 00

2

2

and

dy

dx

2

2

= 0.

dy

dx

x

x

x

2

2

612

01

=− 03

<=,

=,

when a maximum

when a miinimum

==,when a point of inflection















02 x

dy

dx

=−+=− −=312931 30xx xx x x= 1 = 3

2

()() when or

dy

dx

dy

dx

=> 00

2

2

and

dy

dx

dy

dx

=< 00

2

2

and