Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Regarding functions how can simplifying a function change it's domain

  1. Apr 4, 2012 #1
    for example
    if [itex]f(x) = \frac{x^{2} - 1}{x+1}[/itex]

    they by factoring:

    [itex]f(x) = \frac{x^{2} - 1}{x+1}
    = \frac{(x + 1)(x - 1)}{x+1}
    = x - 1 [/itex]

    thus the simplified version is [itex]x - 1[/itex]

    let's say [itex]g(x) =[/itex] simplified [itex] f(x)[/itex]

    [itex]g(x) = x - 1 [/itex]

    meaning [itex]g(x) = f(x)[/itex]

    but

    [itex]g(-1) ≠ f(-1)[/itex]

    am I missing something?
     
    Last edited: Apr 4, 2012
  2. jcsd
  3. Apr 4, 2012 #2
    ##f(x) = \frac {x^2-1}{x+1}##
    ##f(0) = \frac {0^2-1}{0+1}##
    ##f(0) = -1##

    ##g(x) = x-1##
    ##g(0) = 0-1##
    ##g(0) =-1##
    The two functions act the same except at x = -1 at which ##f(x)## has a hole at -1.
     
  4. Apr 4, 2012 #3
    Oops Sorry I meant to write [itex]g(-1) ≠ f(-1) [/itex]
     
  5. Apr 5, 2012 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    Symbolic manipulations in math are an abbreviation for thought, not a substitute for it. If you "simplify" an expression without thinking about it, you can make a faulty deduction. The correct deduction is

    [itex] \frac{(x+1) (x-1)}{x+1} = x-1 [/itex] except when [itex] x = -1 [/itex].

    There may be types of math where you only manipulate symbols and never use ordinary language, but algebra is not one of them.
     
  6. Apr 5, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What you are "missing" is that this is not true. What is true is that f(x)= g(x) for all x except x= -1. f and g are NOT the same function.

     
  7. Apr 5, 2012 #6
    Is there an easy way to spot a problem like this, or is it something you just get better at over time?
     
  8. Apr 5, 2012 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You should have learned long ago that you cannot divide by 0 so that the division in
    [tex]\frac{(x- 1)(x+1)}{x+1}= x- 1[/tex]
    is only valid when [itex]x+ 1\ne 0[/itex] which is only true when [itex]x\ne -1[/itex].
     
  9. Apr 5, 2012 #8

    Stephen Tashi

    User Avatar
    Science Advisor

    There are common traps, like expessions that "don't work" for some values of variables because they involve division by zero or square roots of negative numbers etc.

    You'll get better at spotting the common traps over time if you become suspicious of any form of calculation. Don't trust. Don't believe.
     
  10. Apr 8, 2012 #9
    The dividing by zero makes a lot of sense, I was surprised to see it still apllied even when the numerator would also equal 0 hence 0/0. I will certainly be mindful of this and irrational numbers in the future, thanks for the tip!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Regarding functions how can simplifying a function change it's domain
  1. Domain of a function (Replies: 12)

  2. Domain of a function (Replies: 7)

Loading...