Calculus: Analytic Geometry and Calculus, with Vectors

296 Functions, graphs, and numbers

put plus signs above intervals over which f'(x) > 0 and the graph of f
has positive slope, and (iii) put minus signs above intervals over which
f'(x) < 0 and the graph of f has negative slope. Information about f
can then be obtained with the aid of the two following theorems.
Theorem 5.26 If a < xo < b and if f'(xo) exists, then f cannot have a
maximum or a minimum over the interval a < x < b at xo unless f'(xo) = 0.
Theorem 5.27 If f is continuous over an interval ao < x S be and
f'(x) > 0 when ao < x < bo, then f is increasing over the interval

ao<x<bo.

If f is continuous over an interval ao 5 x < be and f(x) < 0 when

ao<x<bo

then f is decreasing over the interval ao < x < bo.

The second of these theorems is much more forthright and potent than

the first. It will be proved in Section 5.5. The first theorem says that if

a < xo < b and f'(xo) > 0 or f'(xo) < 0, then f cannot have even a local

maximum or a local minimum at xo. To prove this, we suppose first

that a < xo < b and f'(xo) = p, where p is a positive number. Then

lim

Axe + Ax) - f(xo)

= p

AX-0 Ax

and we can choose a positive number S such that

and

a < xo - 3 < xo + 3 < b

Axe + Ax) - Axe )

Ax > 2p

whenever 0 < IAxj < S. If 0 < Ax < 6, then the denominator of the

above quotient is positive, so the numerator must also be positive and

f(xo) < f (xo + Ax). If - S < Ax < 0, then the denominator of the dif-

ference quotient is negative, so the numerator must also be negative and

f(xo + Ax) < f(xo). Thus if xo - 6 < xl < xo < x2 < xo + S, then

f(x1) < f(xo) < f(x2), so f cannot have either a local maximum or a local

minimum at xo. In case a < xo < b and f'(xo) < 0, a similar argument

shows existence of a number 6 such that if xo - 6 < x1 < xo < x2 <

xo + 6, then f (xl) > f (xo) > f (X2) and f cannot have a local maximum or

a local minimum at xo. This completes the proof of Theorem 5.26.

It is quite as important to know what Theorem 5.26 does not imply as it

is to know what the theorem does imply. It does not imply that f has

an extremum (a maximum or a minimum) any place and it does not imply