Calculus: Analytic Geometry and Calculus, with Vectors

306 Functions, graphs, and numbers

To prove the first part of this theorem, let f'(xo) = 0 and let f"(xo) = p,

where p is a positive number. Then

limf'(xo + Ax) - f'(xo) _

AX-0 AX

Let e = p/2. Then there is a positive number S such that f'(xo + Ax)

exists when lAxj < S and

f'(xo + Ax) - f'(xo) p

Ax 2

whenever 0 < jAxi < S. But f'(xo) = 0, and hence

f'(xo +Ax)> 0

Ax

and therefore f'(xo + Ax) and Ax are both positive or both negative when-

ever 0 < jAx! < S. When X o < x < xo + S, we can set x = xo + Ax

and conclude that 0 < Ax < S and f'(x) > 0. When xo - S < x < x0,

we can set x = xo + Ax and conclude that -S < Ax < 0 and f'(x) < 0.

It therefore follows from the last part of Theorem 5.28 that f has a local

minimum at xo. In case f"(xo) < 0, everything is the same except that

some signs are reversed and f has a local maximum at X.

Problems 5.39

1 Sketch a graph of y = 1/x. Calculate dy/dx and d2y/dx' If appro-

priate connections between these things are not immediately clear, there are

only three possibilities: (i) the graph needs repairs or (ii) the formulas for deriva-

tives need repairs or (iii) the text of this section must be studied more carefully.

2 The values of

f(x) =

x

1 + xs

are certainly near 0 when x is near 0 and when jxj is large. Give a full account

of the nature of the graph.

3 Supposing that a, b, and c are constants for which a > 0 and that

f(x) = ax2 + bx + c,

calculate f(x) and f"(x).Show that the only extremum off is a minimum which

is attained when x = -b/2a. Show that the graph off is everywhere bending

upward and that there are no flexpoints.

(^4) Supposing that a, b, c, d are constants for which a > 0 and that
f (x) = axe + bx2 + cx + d,