81 apply the intermediate value theorem to the function f on the interval [−1,0] with γ =0 to conclude that there is c∈[−1,0] such that f (c) =0. The point c is the desired solution to the original equation. ■ Example 3.4.2 We show now that, given n∈N, every positive real number has a positive nth root. Let n∈Nand let a∈Rwith a>0. First observe that (1+a)n ≥1+na>a (see Exercise 1.3.6). Now consider the function f : [0,∞) →Rgiven by f (x) =xn. Since f (0) =0 and f (1+a) >a, it follows from the intermediate value theorem that there is x ∈(0,1+a) such that f (x) =a. That is, xn =a, as desired. (We show later in Example 4.3.1 that such anx is unique.) Now we are going to discuss the continuity of the inverse function. Theorem 3.4.7 Let f : [a,b] →Rbe strictly increasing and continuous on[a,b]. Let c =f (a) and d =f (b). Then f is one-to-one, f ([a,b]) = [c,d], and the inverse function f −1 defined on[c,d] by f −1(f (x)) =x where x ∈[a,b], is a continuous function from[c,d] onto[a,b]. Proof: The first two assertions follow from the monotonicity of f and the intermediate value theorem (see also Corollary 3.4.6). We will prove that f −1 is continuous on[c,d]. Fix any y 0 ∈[c,d] and fix any sequence {yn}in[c,d] that converges to y0. Let x0 ∈[a,b] andxn ∈[a,b] be such that f (x0) =y0 and f (xn) =yn for every n. Then f −1(y 0) =x0 and f − 1(y n) =xn for every n. Suppose by contradiction that {xn} does not converge tox0. Then there exist ε0 >0 and a subsequence {xnk}of {xn}such that |xnk −x0| ≥ε0 for every k. (3.5) Since the sequence {xnk}is bounded, it has a further subsequence that converges to some x1 ∈[a,b]. To simplify the notation, we will again call the new subsequence {xnk}. Taking limits in (3.5), we get |x1 −x0| ≥ε0 >0. (3.6) On the other hand, by the continuity of f , {f (xnk)}converges to f (x1). Since f (xnk) =ynk →y0 as k →∞, it follows that f (x1) =y0 =f (x0). This implies x1 =x0, which contradicts (3.6). □ An analogous theorem can be proved for strictly decreasing functions. Remark 3.4.2 A similar result holds if the domain of f is the open interval (a,b) with some additional considerations. If f : (a,b) →Ris increasing and bounded, following the argument in Theorem 3.2.4 we can show that both limx→a+ f (x) =c and limx→b− f (x) =d exist in R(see Exercise 3.2.11). Using the intermediate value theorem we obtain that f ((a,b)) = (c,d). We can now proceed as in the previous theorem to show that f has a continuous inverse from(c,d) to(a,b). If f : (a,b) →R is increasing, continuous, bounded below, but not bounded above, then limx→a+ f (x) =c ∈R, but limx→b− f (x) =∞ (again see Exercise 3.2.11). In this case we can show using the intermediate value theorem that f ((a,b)) = (c,∞) and we can proceed as above to prove that f has a continuous inverse from(c,∞) to(a,b). The other possibilities lead to similar results.
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