1. One-to-one Or onto?

Hi all,

can anyone help to solve this?

Let f: R -> R defined as f(x) = 9X2. Determine whether or not the function is one-to-one and onto.

2. Originally Posted by cu4mail
Hi all,

can anyone help to solve this?

Let f: R -> R defined as f(x) = 9X2. Determine whether or not the function is one-to-one and onto.

it is not 1-1 nor onto..

not 1-1:
by def:
$if f(x) = f(y) \implies x = y$
take f(x) = 9, then x = 1 or x = -1.
therefore not 1-1.

not onto:
$f: R \rightarrow R$
for every element in the range, there must be an element in domain..
take -1 in range, does there an element in the domain that will give you -1 when mapped using your function?

3. what r the basic rules to identify one-to-one or onto? Just want to understand the methodoligy.

the actual statement was as :
$f: R \rightarrow R$ defined as $f (x) = 9 x^2$

Thanks

4. Originally Posted by cu4mail
what r the basic rules to identify one-to-one or onto? Just want to understand the methodoligy.

the actual statement was as :
$f: R \rightarrow R$ defined as $f (x) = 9 x^2$

Thanks
the definitions themselves:
Def:
A function $f:R \rightarrow R$ is said to be one-to-one if $f(x)=f(y)$, then $x=y$.

okay, let's take your function $f(x)=9x^2$ as an example.
so, suppose $f(x)=f(y)$
then $9x^2 = 9y^2 \implies x^2=y^2 \implies x=|y|$
here, x is either y or -y which contradicts the definition. hence, $f(x)=9x^2$ is not 1-1.

another explanation is that, any value in the range corresponds to a unique value in the domain..

Def:
A function $f:X \rightarrow Y$ is said to be onto Y if the range of f is Y.

in your function, $f:R \rightarrow R$ defined by $f(x)=9x^2$, the range is $[0, +\infty)$ and not R, therefore, f is not onto R.

here, it says that for all the values in the range must corresponds to a value in the domain.