Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/192

Since $f'(a)\neq 0$, and $f'(x)$ is assumed as continuous, h may be chosen so small that $f'(x)$ will have the same sign as $f'(a)$ for all values of x in the interval [a - h, a + h]. Therefore $f'(x_{1})$ has the same sign as $f'(a)$ (Chap. III). But x - a changes sign according as x is less or greater than a. Therefore, from (C), the difference

$f(x)-f(a)$

will also change sign, and, by (A) and (B), $f(a)$ will be neither a maximum nor a minimum. This result agrees with the discussion in § 82, where it was shown that for all values of x for which $f(x)$ is a maximum or a minimum, the first derivative $f'(x)$ must vanish.

II. Let $f'(a)=0$, and $f''(a)\neq 0$.

From (C), § 107, replacing b by x and transposing $f(a)$,

(D)

$f(x)-f(a)={\frac {(x-a)^{2}}{2!}}f''(x_{2}).$

$a<x_{2}<x$

Since $f''(a)\neq 0$, and $f''(x)$ is assumed as continuous, we may choose our interval [a - h, a + h] so small that $f''(x_{2})$ will have the same sign as $f''(a)$ (Chap. III). Also $(x-a)^{2}$ does not change sign. Therefore the second member of (D) will not change sign, and the difference

$f(x)-f(a)$

will have the same sign for all values of x in the interval [a - h, a + h], and, moreover, this sign will be the same as the sign of $f''(a)$. It therefore follows from our definitions (A) and (B) that

(E)	$f(a)$ is a maximum if $f'(a)=0$ and $f''(a)$ = a negative number;
(F)	$f(a)$ is a minimum if $f'(a)=0$ and $f''(a)$ = a positive number.

These conditions are the same as (21) and (22), §84.

III. Let $f'(a)=f''(a)=0$, and $f'''(a)\neq 0.$

From (D), §107, replacing b by x and transposing $f(a)$,

(G)

$f(x)-f(a)={\frac {1}{3!}}(x-a)^{3}f'''(x_{3}).$

$a<x_{3}<x$

As before, $f'''(x_{3})$ will have the same sign as $f'''(a)$. But $(x-a)^{3}$ changes its sign from - to + as x increases through a. Therefore the difference

$f(x)-f(a)$

must change sign, and $f(a)$ is neither a maximum nor a minimum.