Continuity - CONTINUITY Problem 17.4 Prove that the...

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CONTINUITY Problem 17.4: Prove that the function x is continuous on its domain [0 , ) . Proof. Given ± > 0, we need to ﬁnd a δ > 0 such that | x - x o | < δ implies | f ( x ) - f ( x o ) | < ±. | f ( x ) - f ( x o ) | = | x - x o | = ± ± ± ± x - x o x + x o ± ± ± ± < | x - x o | x o < δ x o . ( ? ) We want | f ( x ) - f ( x o ) | < ± . The only problem that we might run into is if x o = 0. So if we have two cases: Case 1: x o > 0. We set δ = ± x o and | f ( x ) - f ( x o ) | < ± from ( * ). Case 2: x o = 0. This is slightly diﬀerent. Since | x - x o | < δ , and in this case x o = 0, | x | < δ . Then x < δ . Thus if we set set δ = ± 2 , we have | f ( x ) - f ( x o ) | = | x - x o | = | x - 0 | < | x | < δ = ±. Thus, let δ = min { ± x o 2 } , then | f ( x ) - f ( x o ) | < ± for all x o [0 , ). Thus, f is continuous. ± Problem 17.6: A rational function is a function f of the form p/q where p and q are polynomial functions. The domain of f is { x R : q ( x ) 6 = 0 } . Prove that every rational function is continuous. d We state problem 17.5 here without proofs (can easily proved using induction): If m N , then the function f ( x ) = x m is continuous on R . Every polynomial function p ( x ) = a o + a 1 x + ... + a n x n is continuous on R . c Proof. Suppose f = p/q for some polynomial functions p,q . Since both p and q are polynomial functions, they are continuous in R . By theorem 17.4, p/q is continuous provided q ( x ) 6 = 0. Since the domain of f is { x R : q ( x ) 6 = 0 } , f is continuous. ± Problem 17.8: Let f and g be real-valued functions. (a) Show that min( f,g ) = 1 2 ( f + g ) - 1 2 | f - g | . (b) Show that min( f,g ) = - max( - f, - g ). (c) Use (a) or (b) to prove that if f and g are continuous at x o R , then min( f,g ) is continuous at x o . Proof. Note that the real hard part is to come up with max( f,g ) = 1 2 ( f + g ) + 1 2 | f - g | . (a) This should be an easy exercise to verify the following statement is true: min( f,g ) = 1 2 ( f + g ) - 1 2 | f - g | . ( ? ) There are two possibilities here: whether f g or f < g . Recall that | a - b | = ² a - b if a b - ( a - b ) if a < b [email protected] 1

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i. If f g , then | f - g | = f - g and min( f,g ) = g . While 1 2 ( f + g ) - 1 2 | f - g | = 1 2 ( f + g ) - 1 2 ( f - g ) = 1 2 f + 1 2 g - 1 2 f + 1 2 g = g = min( f,g ) so ( ? ) holds. ii. If f < g , then | f - g | = - ( f - g ) = g - f and min( f,g ) = f . While 1 2 ( f + g ) - 1 2 | f - g | = 1 2 ( f + g ) - 1 2 ( g - f ) = 1 2 f + 1 2 g - 1 2 g + 1 2 f = f = min( f,g ) so ( ? ) holds as well. (b) Since we have max( f,g ) = 1 2 ( f + g ) + 1 2 | f - g | . Observe that max( - f, - g ) = 1 2 ( - f + ( - g )) + 1 2 | - f - ( - g ) | = - 1 2 ( f + g ) + 1 2 | - ( f - g ) | = - ± 1 2 ( f + g ) - 1 2 | ( f - g ) | ² = - min( f,g ) Thus, min( f,g ) = - max( - f, - g ) .
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Continuity - CONTINUITY Problem 17.4 Prove that the...

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