Calculus: Analytic Geometry and Calculus, with Vectors

5.4 Theorems about continuous and differentiable functions 317

and is positive at x2, and if f is continuous over the closed interval having
end points at x, and x2, then there must be at least one x3 between x, and x2
for which f(x3) = 0. This implies that the graph of a continuous function

f cannot run from a point (xi,yi) below the x axis to a point (x2,y2) above

the x axis without intersecting the x axis at a point (x3,0) for which x3

lies between xi and x2. Instead of asking whether this result is "obvious,"

we can ask whether it is obvious that a man cannot walk from the Capitol

of South Dakota to the Capitol of North Dakota without stepping upon

the common boundary of the two Dakotas.

Problems 5.49

1 With the text of this section out of sight, try to produce adequate responses

to the following orders; if unsuccessful, study the text again and try again.

(a) Write a full statement of the Dedekind axiom.

(b) Write a theorem which gives precise information about boundedness of con-

tinuous functions.

(c) Write a theorem which gives precise information about extrema of con-

tinuous functions.

(d) Write a full statement of the intermediate-value theorem.

2 Using known properties of the function f for which f(x) = x2, show how

the intermediate-value theorem (Theorem 5.48) can be used to prove that there

is a positive number t for which t2 = 2. Give all of the details, recognizing

that Theorem 5.48 cannot be applied until appropriate values of a and b have

been captured.

3 Modify the work of the preceding problem to prove that there must be

at least one x for which f (x) = 0 when

(a) f(x) = x3 - 7 (b) f(x) = x3 - x - 7

a

(c) f(x) = 1 +x2- 40 (d) f(x) = x - cos x

4 Letting

(1) f(x) = 1+x+x2+x3+x4,

determine whether there are any numbers x for which f(x) < 0. Hint: Use the

fact that f(1) = 5 and

(2)

xb - 1

f(x) =x - 1 (x 0 1).

Show that x5 = 1 only when x = 1, so f(x) is never zero and hence f(x) is never

negative.

5 Let f be defined over the closed interval -1 < x 5 1 by the formulas

f(0) = 0 and f(x) = 1/x2 when -1 5 x 5 1 and x F& 0. Show that there is no

constant M such that Jf(x)1 5 Mwhen -1 5 x S 1. Solution: Suppose, intend-

ing to establish a contradiction of the supposition, that there is a number M