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One-to-one Functions.

  1. Oct 4, 2006 #1
    How do you solve one-to-one functions algebraically?

    Problem: f(x) (3x+4)/5

    What are you suppose to substitue for x?

    I know that if f(a)=f(b), a=b...
     
  2. jcsd
  3. Oct 4, 2006 #2

    Office_Shredder

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    What do you mean solve?

    If f(x) = (3x+4)/5, then you have your function. If you plug in an x, you can solve for f(x) by doing basic math. Given what f(x) is, you can do something like this:

    5*f(x) = 3x + 4

    5f(x) - 4=3x

    (5f(x)-4)/3=x
     
  4. Oct 4, 2006 #3
    It says: Determine algebraically whether the function is one-to-one. How do I do that?
     
  5. Oct 4, 2006 #4

    Office_Shredder

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    Oh, I get it. You want to show that if you plug two distinct values, x1 and x2 into f(x), the values that are returned either are not equal to each other, or x1 and x2 are the same (that's the definition of one to one)
     
  6. Oct 4, 2006 #5
    So do you plug 2 and -2. Their negatives? or what two numbers do you plug in. I plugged in 2 and -2 and I got 2 and -2/5. What am I doing wrong?
     
  7. Oct 4, 2006 #6

    Hurkyl

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    Just do what it says:

    Start by writing down f(a) = f(b).
    Use algebra to derive a = b.

    Proof finished.
     
  8. Oct 4, 2006 #7

    berkeman

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    Can you use calculus? If so, you can show that the slope (derivative) of the function is always positive and bounded. That means that the function cannot double-back on itself to create a second y value for any x value.

    Even if you aren't supposed to use calculus, at least for this problem, the equation is the equation of a straight line, right? y = mx + b


    EDIT -- Ooo. I like Hurkyl's method better!
     
  9. Oct 4, 2006 #8
    what do you mean? like f(a)=3a+4/5 f(b)=3b+4/5?? a will always equal b if you do it that way wouldn't it? No I can't use calculus, I'm in pre cal xD. Can you show me how to do it?
     
  10. Oct 4, 2006 #9

    Hurkyl

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    If you can prove what you just said, then you've proven f is one-to-one. It's that easy.
     
  11. Oct 4, 2006 #10
    What about f(x)=x^2 It's not a one-to-one fucntion even though f(a)=a^2 is equal to f(b)=b^2 a=b in that case but if you put -2 and 2 in f(a)=f(b) but a doesn't equal b.. So I'm confused.
     
  12. Oct 4, 2006 #11
    Can you give me some examples of functions that aren't one-by-one. I mean if you use the same function for x=a and b. aren't they alwasy equal to each other?
     
  13. Oct 4, 2006 #12

    Office_Shredder

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    For x^2, if a^2=b^2, then a=-b means x^2 isn't necessarily one to one
     
  14. Oct 4, 2006 #13

    Hurkyl

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    Why do you think a=b in that case? You've demonstrated that's not always true... so think hard about why you would (incorrectly) believe a=b must be true here.
     
  15. Oct 4, 2006 #14
    Okay I know that -2^2 and 2^2 are both equal to 4.. So that means you can't use a and b.. cause a^2 and b^2 would be a=b if you solve it algebrically. What numbers do I need to substitue for x? How do I know that a=b or a doesn't equal b. Thats what I'm trying to figure out. =P
     
  16. Oct 4, 2006 #15

    Hurkyl

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    Show your work.
     
  17. Oct 4, 2006 #16
    a^2=b^2 because you square root both sides and you get a=b?
     
  18. Oct 4, 2006 #17

    Hurkyl

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    Nope. You get |a| = |b|.
     
  19. Oct 4, 2006 #18
    So basically anything that is to an even power is not a one-to-one function. ok this is weird. lol
     
  20. Oct 5, 2006 #19

    berkeman

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    Just for my peace of mind, could you please post the textbook definition of a one-to-one function? I think that my practical definition of a one-to-one function (any x maps to only one y) may not match what others are asking you to show.
     
  21. Oct 5, 2006 #20
    One-to-one means IF f(x) = f(y) THEN x=y.

    So think about f(x) = x^2. If f(x) = f(y) is it NECESSARILY true that x=y? If not, then f is not 1-1. If so, then f is 1-1.
     
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