254 Chapter 5 • Continuous Functions
Proof. If n = 1 we merely take y = xo.
Suppose n ~ 2 in N, and x 0 > 0 in R Consider the (polynomial) function
p(x) = xn. We know that p(O) = 0. By Theorem 4.4.18, lim p(x) = + oo,
x->+oo
so 3b E (0, +oo) 3 p(b) > x 0. Thus, by the intermediate value theorem, 3
y E (0, b) 3 p(y) = x 0. Uniqueness follows from Lemma 5.3.15. •
EXERCISE SET 5.3l. Suppose f : V(f) -t JR and A ~ V(f), and consider Statements #1 and
#2 in the introductory paragraphs of this section. Prove that these two
statements are equivalent if and only if A is open. [Hint: If A is not open,
try the characteristic function XA·]- Prove that f : [a, b] -t JR is continuous ~ f is continuous on (a, b) and
lim f(x) = f(a) and lim f(x) = f(b).
x->a+ x->b- - Prove Theorem 5.3.4.
- Give an example for each of the following: (Draw a graph to explai n your
reasoning.)
(a) A function that exists on [O, 2] and has a maximum value on [O, 2],
but does not have a minimum value there.
(b) A function that exists on [O, 2] and has a minimum value on [O, 2],
but does not have a maximum value there.
(c) A function that exists on [O, 2] and has neither a minimum value on
[O, 2] nor a maximum value there.
(d) A function that is continuous on [1, +oo) and has a maximum value
there, but no minimum value there.
(e) A function that is continuous on [1, +oo) and has a minimum value
there, but no maximum value there.
(f) A function that is continuous on [1, + oo) and has neither a minimum
value nor a maximum value there.
(g) A function that is continuous on (-1, 1) and has neither a minimum
value nor a maximum value there.
(h) A function that is continuous on (-1,1) and has a minimum value
on (-1, 1), but does not have a maximum value there.
(i) A function that is continuous on (-1, 1) and has a maximum value
on ( -1, 1), but does not have a minimum value there.- Prove the extreme value theorem (Corollary 5.3.7).