80 3.4 Properties of Continuous Functions This set is nonempty since a ∈ A and bounded since A ⊂ [a,b]. Therefore, c = supA exists. Since c = supA, we can fnd a sequence {xn} in A such that xn → c (see Exercise 2.1.10). Applying the comparison theorem for sequences (Theorem 2.1.2), we obtain a ≤ c ≤ b and since 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 fnd x ∈ (c,b) such that f (x) < 0. This is a contradiction because x ∈ A and x > 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] → R be 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: Defne ϕ(x)= f (x) − γ, x ∈ [a,b]. Then ϕ 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] → R be a continuous function. Let m = min{ f (x) : x ∈ [a,b]} and M = 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 equation e x = −x has at least one real solution. We will assume known that the exponential function is continuous on R and that ex < 1 for x < 0. First defne the function f : R → R by 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
RkJQdWJsaXNoZXIy NTc4NTAz