79 For every n∈N, there exists xn ∈[a,b] such that α−1/n<f (xn) ≤α. Since a ≤xn ≤b for every n, the sequence is bounded. Applying the Bolzano-Weierstrass theorem (Theorem 2.4.1), there exists a convergent subsequence, say, {xnk}, and x0 such that lim k→∞ xnk =x0. Note that x0 ∈[a,b] by the Comparison theorem (Theorem 2.1.2). From the continuity of f on [a,b] we have lim k→∞ f (xnk) =f (x0). On the other hand, α− 1 nk <f (xnk) ≤α for every k and applying the Squeeze theorem (Theorem 2.1.3) we get that limk→∞ f (xnk) =α. Since f is a continuous function and limk→∞xnk =x0, we have limk→∞ f (xnk) =f (x0). Therefore f (x0) =α and f (x0) ≥ f (x) for every x ∈[a,b]. Thus, f has an absolute maximum at x0. Let xM=x0. Similarly, considering −f instead of f , we can show the existence of the absolute minimum. The result follows. □ The following result is a basic property of continuous functions that is used in a variety of situations. Lemma 3.4.3 Let f : D→Rbe continuous at c ∈D. Suppose f (c) >0. Then there exists δ >0 such that f (x) >0 for every x ∈Dsuch that |x−c| <δ Proof: Let ε = f (c) 2 >0. By the continuity of f at c, there exists δ >0 such that if x ∈Dand |x−c| <δ, then | f (x)−f (c)| <ε. This implies, in particular, that f (x) >f (c)−ε = f (c) 2 >0. Pick this δ, then the result follows. □ Remark 3.4.1 An analogous result holds if f (c) <0. Theorem 3.4.4 Let f : [a,b] →Rbe a continuous function. Suppose 0 is strictly between f (a) and f (b) (that is, either f (a) <0 < f (b) or f (a) >0 > f (b)). Then there exists c ∈(a,b) such that f (c) =0. Proof: We prove only the case f (a) <0<f (b) (the case f (a) >0>f (b) is completely analogous). Define A={x ∈[a,b] : f (x) ≤0}.
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