# Determine whether it is an onto function

1. May 8, 2010

### jwxie

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. May 8, 2010

### Martin Rattigan

$x^2+x=(x+\frac{1}{2})^2-\frac{1}{4}$, so $g_1(x)>=-\frac{1}{4}$. So, for example, -1 is not in $g_1(\mathbb{R})$.

3. May 8, 2010

### Martin Rattigan

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