80 3.4 Properties of Continuous Functions This set is nonempty since a∈Aand bounded since A⊂[a,b]. Therefore, c=supAexists. Since c =supA, we can find a sequence {xn} in Asuch that xn →c (see Exercise 2.1.10). Applying the comparison theorem for sequences (Theorem 2.1.2), we obtaina≤c≤bandsince f is continuous on [a,b], f (xn) →f (c). Applying the comparison theorem one more time, we conclude that f (c) ≤0. We will prove that f (c)=0 by showing that f (c) <0 leads to a contradiction. Suppose f (c) <0. Then, by Remark 3.4.1, there exists δ >0 such that f (x) <0 for all x ∈[a,b] such that |x−c| <δ. Because c <b (since f (b) >0), we can find x ∈(c,b) such that f (x) <0. This is a contradiction because x ∈Aandx >c. We conclude that f (c)=0. □ Figure 3.2: Illustration of the Intermediate Value Theorem. Theorem 3.4.5 — Intermediate Value Theorem. Let f : [a,b] →Rbe a continuous function. Suppose f (a) <γ <f (b). Then there exists a number c ∈(a,b) such that f (c)=γ. The same conclusion follows if f (a) >γ >f (b). Proof: Define ϕ(x)=f (x)−γ, x ∈[a,b]. Thenϕ is continuous on[a,b]. Moreover, ϕ(a) <0<ϕ(b). By Theorem 3.4.4, there exists c∈(a,b) such that ϕ(c)=0. This is equivalent to f (c)=γ. The proof is now complete. □ Corollary 3.4.6 Let f : [a,b] →Rbe a continuous function. Let m=min{f (x) : x ∈[a,b]} andM=max{f (x) : x ∈[a,b]}. Then for every γ ∈[m,M], there exists c ∈[a,b] such that f (c)=γ. ■ Example 3.4.1 We will use the Intermediate Value Theorem to prove that the equatione x =−x has at least one real solution. We will assume known that the exponential function is continuous on Rand that ex <1 for x <0. First define the function f : R→Rby f (x)=ex +x. Notice that the given equation has a solution x if and only if f (x)=0. Now, the function f is continuous (as the sum of continuous functions). Moreover, note that f (−1)=e−1 +(−1) <1−1=0 and f (0)=1>0. We can now
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