5333-Solns-Assignment7-Sp13 - Solutions Assignment 7 Section 23 3(d f(x = x2 3x 5 on D =(1 3 Dene g on[1 3 by f(x x(1 3 g(x = 1 x=1 13 x=3 Then g(x is

# 5333-Solns-Assignment7-Sp13 - Solutions Assignment 7...

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Solutions, Assignment 7 Section 23. 3. (d) f ( x ) = x 2 + 3 x - 5 on D = (1 , 3). Define g on [1 , 3] by g ( x ) = f ( x ) , x (1 , 3) - 1 , x = 1 13 , x = 3 Then g ( x ) is continuous on [1 , 3] which implies that f is continuous on (1 , 3) by Theorem 23.9. 4. (a) Let epsilon1 > 0. For x, y [0 , 2], | x 3 - y 3 | = | ( x - y )( x 2 + xy + y 2 ) | = | x 2 + xy + y 2 | | x - y | ≤ 12 | x - y | . Choose δ = epsilon1 12 . Then, for x, y [0 . 2] and | x - y | < δ, , it follows that | x 3 - y 3 | = | x 2 + xy + y 2 | | x - y | ≤ 12 | x - y | < 12 epsilon1 12 = epsilon1. Therefore, f ( x ) = x 3 is uniformly continous on [0 , 2]. 5. f ( x ) = x . Let epsilon1 > 0. Since f is continuous on [0 , 1], there is a positive number δ 1 such that | f ( x ) - f ( y ) | < epsilon1/ 2 whenever | x - y | < δ 1 , x, y [0 , 1]. Now consider f on [1 , ). Let x, y [1 , ); suppose x < y : vextendsingle vextendsingle x - y vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle ( x - y ) x + y x + y vextendsingle vextendsingle vextendsingle vextendsingle = 1 x + y | x - y | ≤ 1 2 x | x - y | ≤ 1 2 | x - y | . Let δ 2 = epsilon1 . Then, vextendsingle vextendsingle x - y vextendsingle vextendsingle < epsilon1/ 2 whenever | x - y | < δ 2 , x, y [1 , ). Let δ = min { δ 1 , δ 2 } . Suppose x [0 , 1] and y [1 , ). Then vextendsingle vextendsingle x - y vextendsingle vextendsingle = vextendsingle vextendsingle x - 1 + 1 - y vextendsingle vextendsingle vextendsingle vextendsingle

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