F : N -> N defined by f(x) = x^3 - 1

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SUMMARY

The discussion focuses on the properties of three mathematical functions: f(x) = x^3 - 1, g(x) = 2x + 1, and h(x) = x(x + 3)(x - 3). It addresses whether these functions are one-to-one and bijective, and it seeks to find inverse functions for those that are bijections. The function f(x) = (3x - 1)/(x - 3) is noted as not bijective unless its domain and codomain are suitably restricted, prompting further exploration of these concepts.

PREREQUISITES
  • Understanding of function definitions and notation (e.g., f : N --> N)
  • Knowledge of one-to-one and bijective functions
  • Familiarity with inverse functions and their calculations
  • Basic concepts of domain and codomain in mathematics
NEXT STEPS
  • Study the properties of one-to-one functions in detail
  • Learn about bijective functions and their significance in mathematics
  • Explore how to find inverse functions for various types of functions
  • Investigate domain and codomain restrictions to achieve bijectivity
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Students of mathematics, educators teaching function properties, and anyone interested in understanding the concepts of one-to-one and bijective functions.

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f : N --> N defined by f(x) = x^3 - 1

Hi all, I need answers and EXPLANATION to the following problems: (Please Help!)

(i) f : N --> N defined by f(x) = x^3 - 1
(ii) g : Z --> Z defined by g(x) = 2x + 1
(iii) h : R --> R defined by h(x) = x(x + 3)(x - 3)

(***note that N,Z,R stands for natural #, Integer, and Real # respectively..)

(a) Which of the functions are one-to-one?(b) Which of the functions are bijections?(c) For those that are bijections find the inverse function.--------------------------------------------------------------------------------------------------------------
Here's the other one:

The function f : R --> R defined by f(x) = (3x - 1)/(x - 3) is not bijective however by suitably restricting the domain and codomain the function can be made to be bijective.

(a)State the domain and codomain that will make the function bijective.

What's a domain? codomain?

(b) Find the inverse of the bijective function.

(I can still remember a bit of inverse function.. i think.. well ill give it a try anyway)

f(x) = (3x - 1)/x - 3)
x = (3y - 1)/y - 3) "replace x with y"
x(y - 3) = 1(3y - 1)) "Cross multiplication"
xy - 3x[i/] = 3y - 1 "Will minus both sides with 3y"
xy - 3x - 3y = -1 "Will add both sides with 3x"
xy - 3y = -1 + 3x
y(x - 3) = -1 + 3x "Factor out y"
y = (-1 + 3x)/(x - 3) "Divide both sides with (x - 3)"

Is that correct??
 
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This site is not an answer service! We will be happy to help you work through the problem, but this requires you to show what you've done on the problem so far (even if it's just pondering what you can do, or what needs to be done).


I think you stated something incorrectly as well -- the expression you wrote as a definition for f doesn't define a function from R to R.
 

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