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 — xr, <//(#) = 3a; 2 , which 
vanishes when x = 0. But x = 0 does not give either a maximum 
or a minimum of ¿c 3 , as is obvious from the form of the graph 
of a? (§ 13, Fig. 11). 
But there will certainly he a maximum for x = % if (£) = 0, 
(f)' (x) >0 for all values of x less than hut near to x, and <f>' (x) < 0 
for all values of x greater than hut 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 (£— 8, %) throughout 
which cf)(x) increases with x, and an interval (£, £ + 8) throughout 
which it decreases as x increases : and obviously this ensures that 
(p(f) shall be a maximum. 
This result may also be stated thus—if the sign of <£' (x) 
changes at x = £ from positive to negative, then x = £ gives 
a maximum of </>(#): and if the sign of ffi(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 (f)(x) has a second derivative 4>'\x): this of course does not 
follow from the existence of ffi(x), any more than the existence of 
cp'(x) follows from that of <£(&•). But in such cases as we are 
likely to meet with at present the condition is generally satisfied. 
Theorem D. If <£'(£) = 0 and <p' (f) 0, <p(x) has a maximum 
or minimum for x=£: a maximum if <£"(£) <0, a minimum if 
</>"(£) >o. 
Suppose, e.g., $"(!;) <0. Then cf)'(x) is decreasing near &• = £, 
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-d{x 2 ), 
where the square root is to be taken positive. The graph of this function is 
shown in Fig. 48. Here <£(- l) — 4>(1) = 0: but 4>’{x) (as is evident from the 
figure) is equal to +1 if x is negative and to — 1 if x is positive, and never