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MAS111_09_assignment_13_1_hints

# MAS111_09_assignment_13_1_hints - MAS 111 FOUNDATION OF...

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MAS 111 FOUNDATION OF MATHEMATICS PRACTICE EXERCISES FUNCTIONS — Hints 1. Does the formula f ( x ) = 1 x 2 - 2 define a function f : R -→ R ? HInt : No. f ( 2) and f ( - 2) are not defined. 2. For each of the following functions, determine whether it is one-to-one and determine its range. (a) f : Z -→ Z , f ( x ) = 2 x + 1; (b) g : Q -→ Q , g ( x ) = 2 x + 1; (c) h : Z -→ Z , h ( x ) = x 3 - x ; (d) k : R -→ R , k ( x ) = e x ; (e) l : [ - π/ 2 , π/ 2] -→ R , l ( x ) = sin x ; (f) j : [0 , π ] -→ R , j ( x ) = sin x . Hint : (a) f : Z -→ Z , f ( x ) = 2 x + 1 is one-to-one, and its range is the set of odd integers; (b) g : Q -→ Q , g ( x ) = 2 x + 1 is one-to-one, and its range is Q ; (c) h : Z -→ Z , h ( x ) = x 3 - x is not one-to-one, because h (0) = h (1) = h ( - 1) = 0. Its range is { n Z : n is the product of three consecutive integers } ; (d) k : R -→ R , k ( x ) = e x is one-to-one, and its range is R + , the set of positive real numbers; (e) l : [ - π/ 2 , π/ 2] -→ R , l ( x ) = sin x is one-to-one, and its range is [ - 1 , 1]; (f) j : [0 , π ] -→ R , j ( x ) = sin x is not one-to-one, because j (0) = j ( π ) = 0 (you may find some other witnesses), and its range is [0 , 1]. 1

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3. For each of the following functions g : R -→ R , determine whether the function is one-to-one and whether it is onto. If the function is not onto, determine the range g ( R ). (a) g ( x ) = x + 7; (b) g ( x ) = 2 x - 3; (c) g ( x ) = - x + 5; (d) g ( x ) = x 2 ; (e) g ( x ) = x 2 + x ; (f) g ( x ) = x 3 . Hint : Similar to question 2. 4. Let A = { x R : - 1 < x < 1 } . Show that the function R -→ A defined by f ( x ) = x 1 + | x | is a one-to-one and onto function. Hint : We first show that f is a function from R to A . For any x R , | f ( x ) | = | x 1 + | x | | = | x | 1 + | x | < 1 because | x | < 1 + | x | is always true. Therefore, f ( x ) ( - 1 , 1). Obviously, every x in R is mapped to a unique number, x 1+ | x | , in ( - 1 , 1). Therefore, f is a function.
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