Unformatted text preview: h ( x ) = f ( x ) − g ( x ). By Theorem 73, h is continuous. Also h ( a ) = f ( a ) − g ( a ) ≤ 0 since f ( a ) ≤ g ( a ), and h ( b ) = f ( b ) − g ( b ) ≥ 0 since f ( b ) ≥ g ( b ). If h ( a ) = 0, then we have f ( a ) = g ( a ). Likewise, if h ( b ) = 0 we have f ( b ) = g ( b ). If h ( a ) , h ( b ) n = 0, then by Lemma 4, there is some c ∈ ( a, b ) such that h ( c ) = 0 which implies f ( c ) = g ( c ). ˜ 3. Suppose that f is a realvalued continuous function on R and that f ( a ) f ( b ) < 0 for some a, b ∈ R . Prove that there exists c between a and b such that f ( c ) = 0. Proof: If f ( a ) f ( b ) < 0, then we have the following two cases: Either f ( a ) > 0 and f ( b ) < 0 or f ( a ) < and f ( b ) > 0. In either case, we can apply Lemma 4 to ±nd a value c ∈ ( a, b ) such that f ( c ) = 0. ˜...
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This document was uploaded on 10/01/2009.
 Spring '09
 Math

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