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Determine whether it is an onto function

  1. May 8, 2010 #1
    If it is not an onto function, provide the range.

    I have problems with these two

    g: R -> R
    g(x) = x^2 + x
    g(x) = x^3

    The book said g1 is neither onto nor 1-1, while g2 is both

    I see how x^2 + x is not `1-1 function. But how come it is not onto? I mean R is real number, can be 343.3434534564656757 or 67.23123132 or anything that is real number. How can we quickly show that x^2 + x is not onto? There are -inf, inf real numbers, so how can we say there is a particular g(x) left out?
    One way I see is if x^2 + x = 0 and we get x = sqrt(-x) which is an i num

    I see x^3 is a 1-1 because x1 = x2. in a similar way as in the case of g1, how can we be so sure that all real numbers can be produce?

    In the book, things like x^3 its surjectiveness can be proven as following
    x^3 and if r is any real number in codomain of f, then real number cubic root of r is in the domain of f and since f(cubic root of r) = (cubic root of r)^3 = r, so f = R = range of f, so it is onto (copied from book)

    Thanks
     
  2. jcsd
  3. May 8, 2010 #2
    [itex]x^2+x=(x+\frac{1}{2})^2-\frac{1}{4}[/itex], so [itex]g_1(x)>=-\frac{1}{4}[/itex]. So, for example, -1 is not in [itex]g_1(\mathbb{R})[/itex].
     
  4. May 8, 2010 #3
    As for the second you if y is any real number then we can find x0 and x1 with x03<y<x13 (not difficult to prove - I'll leave the details to you).

    The existence of some real number x such that x3=y then follows from the intermediate value theorem.

    You probably need to find a proof of the IVP - try wiki.

    PS. You also need to prove x3 is continuous.
     
    Last edited: May 8, 2010
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