23b_hw2_solutions

23b_hw2_solutions - Math 23b Homework 2 Solutions 1. In...

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Math 23b Homework 2 Solutions 1. In each part, the rules are given by well-defined functions on certain domains whose union is R , so we just need to check that these functions agree wherever their corre- sponding domains overlap. (a) For 2 < x < 4, x > x - 1 > 0, so | x - 1 | = x - 1 = | x | - 1, so the rules define a function on R . (b) Note that - 1 < 0 < 2 so 0 lies in both domains. Since | 0 - 1 | = 1 6 = - 1 = | 0 |- 1, this implies the rules do not define a function. (c) There is no overlap in the domains, so the rules define a function on R . (d) Expanding the numerator of (( x +3) 2 - 9) /x gives ( x +3) 2 - 9 = x 2 +6 x +9 - 9 = x ( x + 6). Hence, for x 6 = 0, we have (( x + 3) 2 - 9) /x = x + 6, so the functions agee on the intersection of their domains, so the rules define a function on R . (e) The first two domains both contain the interval [2 , 4], but for x 2 > 0, the square root theorem gives x 2 = x , so the rules define a function on R . 2. Let f and g be real-valued functions on R . (a) Claim: If f and g are bounded, then f + g is also bounded. Proof. By assumption, there exist constants M and N such that | f ( x ) | ≤ M and | g ( x ) | ≤ N for all x R , so by the triangle inequality, we have for each
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23b_hw2_solutions - Math 23b Homework 2 Solutions 1. In...

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