2.1 DIFFERENTIATION
gradient of the function is zero but the function rises in the positivex-direction
and falls in the negativex-direction is also a stationary point of inflection.
The above distinction between the three types of stationary point has beenmade rather descriptively. However, it is possible to define and distinguish sta-
tionary points mathematically. From their definition as points of zero gradient,
all stationary points must be characterised bydf/dx= 0. In the case of the
minimum,B, the slope, i.e.df/dx, changes from negative atAto positive atC
through zero atB. Thusdf/dxis increasing and so the second derivatived^2 f/dx^2
must be positive. Conversely, at the maximum,Q, we must have thatd^2 f/dx^2 is
negative.
It is less obvious, but intuitively reasonable, that atS,d^2 f/dx^2 is zero. This maybe inferred from the following observations. To the left ofSthe curve is concave
upwards so thatdf/dxis increasing withxand henced^2 f/dx^2 >0. To the right
ofS, however, the curve is concave downwards so thatdf/dxis decreasing with
xand henced^2 f/dx^2 <0.
In summary, at a stationary pointdf/dx= 0 and(i) for a minimum,d^2 f/dx^2 >0,(ii) for a maximum,d^2 f/dx^2 <0,(iii) for a stationary point of inflection,d^2 f/dx^2 = 0 andd^2 f/dx^2 changes sign
through the point.In case (iii), a stationary point of inflection, in order thatd^2 f/dx^2 changes signthrough the point we normally required^3 f/dx^3 = 0 at that point. This simple
rule can fail for some functions, however, and in general if the first non-vanishing
derivative off(x) at the stationary point isf(n)then ifnis even the point is a
maximum or minimum and ifnis odd the point is a stationary point of inflection.
This may be seen from the Taylor expansion (see equation (4.17)) of the function
about the stationary point, but it is not proved here.
Find the positions and natures of the stationary points of the function
f(x)=2x^3 − 3 x^2 − 36 x+2.The first criterion for a stationary point is thatdf/dx= 0, and hence we set
df
dx=6x^2 − 6 x−36 = 0,from which we obtain
(x−3)(x+2)=0.Hence the stationary points are atx=3andx=−2. To determine the nature of the
stationary point we must evaluated^2 f/dx^2 :
d^2 f
dx^2=12x− 6.