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One to one functions/range -calc II

  1. Sep 5, 2007 #1
    One to one functions/range --calc II

    In calc II, we are learning about inverse functions, So I have two questions:

    1. Is there any way to find out if a function is one to one without looking at the graph?

    2. How can I determine the range of a function based on the equation alone (without referring to the graph)?

    -Lastly, there is this example on the book that says f(x) = 1/4x^3+x-1 and asks "what is f^-1(x) when x= 3?" The explanation just says that f(x)=3 when x=2 so f^-1(3)=2. But how did they find f(2)=3?? I tried doing that but came accross 3 solutions--is looking at the graph necessary to know this?

    Thank you.
  2. jcsd
  3. Sep 5, 2007 #2


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    1) Yes,it depends on what the function is,normally polynomials are restricted to some domain.

    2)Well I could give you an example where you don't have to look at the graph..but for polynomials I am not sure if you can just write down the range and so forth

    [tex]f(x) = \frac{1}{4}x^3+x-1[/tex] what is f^-1(x) when x= 3

    so what you want to find is [tex]f^{-1}(3)[/tex] if we let [tex]f^{-1}(3)=a[/tex] where a is the answer you want to find.. then you can say that [tex]3=f(a)[/tex] then just simply solve [tex]\frac{1}{4}a^3+a-1=0[/tex] which will give you [tex](a-2)g(a)=0[/tex]

    now check if the [tex]g(a)[/tex] has any real solutions, if not then [tex]a=2[/tex] is your only answer
  4. Sep 5, 2007 #3


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    1) Yes. For example suppose f(x) = 3x + 2 where the domain and range are the real numbers. We can easily prove this is one to one... The method is to assume you have two x's... x1 and x2, such that:

    f(x1) = f(x2)
    => 3x1 + 2 = 3x2 + 2
    => 3x1 = 3x2
    => x1 = x2

    So this means that the function is one to one... the only way for f(x1) to be the same as f(x2) is if x1=x2... ie: you can't have two different x's that have the same f(x).

    To prove that a function is not one to one, you just need one counterexample... for example

    f(x) = x^2 (domain is the reals)

    f(1) = f(-1), so this shows that the function isn't one to one.
  5. Sep 5, 2007 #4


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    1) Rotate the co-ordinate axis using the normal rotational transformation formulae and see if you still get a functional description.

    2)Check the absolute maximum and absolute minimum of the function.
  6. Sep 5, 2007 #5
    how did you come up with [tex](a-2)g(a)=0[/tex]?
  7. Sep 5, 2007 #6


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    ah my bad...solve [tex]|frac{1}{4}a^3+a-4=0[/tex]
  8. Sep 5, 2007 #7
    ok, if I solve that: a(1/4a^2 +1)= 3...I still don't get it. If you or anyone can go through the steps, I would appreciate it very much.
  9. Sep 5, 2007 #8


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    ok uhmm let me help:



    Now (a-2) is a factor of this equation so now applying synthetic division

    [tex]a^3+4a-16\equiv (a-2)(a^2+Pa+8)[/tex] where P is constant
    Equating the terms of [tex]a^2[/tex] on both sides of the equation
    and so [tex]0=-2+P \Rightarrow P=2[/tex]

    so you now get [tex](a-2)(a^2-2a+8)=0[/tex]
    [tex](a^2-2a+8)[/tex] has no real roots as [tex](-2)^2<4(1)(8)[/tex]

    and so [tex]a=2[/tex] is the answer
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