A course of pure mathematics, by G.H. Hardy.

102-105]

DERIVATIVES AND INTEGRALS

207

That this condition is not sufficient is evident from a glance
at the point C in the figure. Thus if y =x, b)'(x)=3x2, which
vanishes when x = 0. But x = 0 does not give either a maximum
or a minimum of x3, as is obvious from the form of the graph
of x3 (~ 13, Fig. 11).
But there will certainly be a maximum for x= 4 if /' (4:)= 0,
' (x) > 0 for all values of x less than but near to x, and ' (x) < 0
for all values of x greater than but near to x: and if the signs of
these two inequalities are reversed there will certainly be a
minimum.
For then we can determine an interval (I-8, 4) throughout
which b (x) increases with x, and an interval (I, + 8) throughout
which it decreases as x increases: and obviously this ensures that
0b(4) shall be a maximum.
This result may also be stated thus-if the sign of b'(x)
changes at x from positive to negative, then x = gives
a maximum of ((x): and if the sign of q'(x) changes in the
opposite sense, then x = gives a minimum.
104. There is another way of stating the conditions for a
maximum or minimum which is often useful. Let us assume
that +(x) has a second derivative +"(x): this of course does not
follow from the existence of +'(x), any more than the existence of
+'(x) follows from that of (x). But in such cases as we are
likely to meet with at present the condition is generally satisfied.
THEOREM D. If 0'(f) = 0 and ' (I) =j 0, (b(x) has a maximum
or minimum for x=: a maximum if +b"(t) < 0, a mi imum if
qi (4) >o.
Suppose, e.g., +"(:) < 0. Then +'(x) is decreasing near x= I:,
and so its sign changes from positive to negative. Thus x=
gives a maximum.
105. In what has preceded (apart from the last paragraph) we have
assumed simply that (x) has a derivative for all values of x in the interval
under consideration. If this condition is not fulfilled the theorems cease to
be true. Thus Theorem B fails in the case of the function
y=l - /(X2),
where the square root is to be taken positive. The graph of this function is
shown in Fig. 48. Here /(- 1)=((1)=0: but 0'(x) (as is evident from the
figure) is equal to + 1 if x is negative and to -1 if x is positive, and never