Calculus: Analytic Geometry and Calculus, with Vectors

5.2 Trends, maxima, and minima 297

that f has an extremum at xo. It does imply that if f hasan extremum at

xo, then xo must be either

(i) one of the end pointsa and b or

(ii) such that f'(xo) doesnot exist or

(iii) such that f'(xo)= 0.

The points xo in these three categories are therefore the onlyones that

need be examined when we are seeking extrema of fover the interval

a 5 x < b. This information, meager as it is, is often helpful. Figure

5.271 may help us to understand it. The following theorems, which are

a c d o f g b x

Figure 5.271

easily interpreted in terms of Figure 5.25 and whichare easy consequences

of Theorem 5.27, give all the information we need to solvemany problems.

Theorem 5.28 (maximum) If f is continuous over a S x 5 b, if

a < xo < b, and if there is a positive number h such that f(x) > 0 when

xo - h < x < xo and f'(x) < 0 when xo < x < xo + h, then f has a local

maximum (which may be a global maximum) atxo.

Theorem 5.281 (minimum) If f is continuous over a <- x 5 b, if

a < xo < b, and if there is a positive number h such that f'(x) < 0 when

xo - h < x < xo and f' (x) > 0 when xo < x < xo + h, then f has a local

minimum (which may be a global minimum) atxo.

Theorem 5.28 says, in slightly different words, that ifwe travel a

smooth road in such a way that we go uphill from 8:58 A.M. to 9:00A.M.

and go downhill from 9:00 A.M. to 9:02 A.M., we are atopa hill (but not

necessarily atop the highest mountain) at 9:00 A.M. Theorem 5.281 has a

similar interpretation.

Problems 5.29

(^1) Letting f be defined over El by the formula f(x) = x2 - 2x + 3, show that
f'(x)=2x-2=2(x-1).
Observe that f'(1) = 0, and then make the more profound observation that
f'(x) < 0 and f is decreasing when x < 1 and that f(x) > 0 and f is increasing
when x > 1. Show that f(l) = 2 and use the information to sketch a graph of
y = f (x). Give all of the facts involving extrema (maxima and minima) of f.