{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MAS111_09_assignment_13_1_hints - MAS 111 FOUNDATION OF...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern