HW16Solutions - h ( x ) = f ( x ) g ( x ). By Theorem 73, h...

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Math 3210-3 HW 16 Solutions Properties of Continuous Functions 1. Show that 2 x = 3 x for some x (0 , 1). Proof: First I claim that f ( x ) = 2 x is a continuous function on [0 , 1]. Let’s assume e x and ln x are continuous functions on R . Then consider y = ln 2 x = x ln 2. This is a polynomial function since it is simply a constant (ln 2) times x . We already proved polynomial functions are continuous, so y is continuous. Now notice that 2 x = e ln 2 x = e x ln 2 = e y . Since the composition of continuous functions is continuous, 2 x is continuous. Now for the proof of the problem. Let g ( x ) : R R be de±ned as g ( x ) = 2 x 3 x . We have already shown that polynomial functions are continuous, so By Theorem 73, g is continuous. Also notice that g (0) = 1 > 0 and g (1) = 2 3 = 1, so by Lemma 4, there is some c (0 , 1) such that g ( c ) = 0. This implies that 2 c = 3 c for some c (0 , 1). ˜ 2. Suppose that f : [ a, b ] R and g : [ a, b ] R are continuous functions such that f ( a ) g ( a ) and f ( b ) g ( b ). Prove that f ( c ) = g ( c ) for some c [ a, b ]. Proof: Let h ( x ) : [ a, b ] R be de±ned by
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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 real-valued 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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